Is the theory of the category of topological spaces computable?How to state the Axioms of Category Theory in...

What is the difference between ashamed and shamed?

What's a good word to describe a public place that looks like it wouldn't be rough?

Is it a fallacy if someone claims they need an explanation for every word of your argument to the point where they don't understand common terms?

Auto Insert date into Notepad

Sometimes a banana is just a banana

Eww, those bytes are gross

Can the Grease spell force multiple saves?

Do my Windows system binaries contain sensitive information?

How would an AI self awareness kill switch work?

How to use a mathematical expression as xticklable

Can the Count of Monte Cristo's calculation of poison dosage be explained?

How to acknowledge an embarrassing job interview, now that I work directly with the interviewer?

Meaning of すきっとした

Charged enclosed by the sphere

Incompressible fluid definition

Prove that every even perfect number is a triangular number.

Can a hotel cancel a confirmed reservation?

If a druid in Wild Shape swallows a creature whole, then turns back to her normal form, what happens?

Six real numbers so that product of any five is the sixth one

I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?

How do we edit a novel that's written by several people?

It took me a lot of time to make this, pls like. (YouTube Comments #1)

Has the Isbell–Freyd criterion ever been used to check that a category is concretisable?

Can I retract my name from an already published manuscript?



Is the theory of the category of topological spaces computable?


How to state the Axioms of Category Theory in Predicate Logic?Properties of the Category of topological spaces with $n$ basepoints.Developing category theory inside ETCSConcrete category of topological spaces over preordered sets: what are the initial morphisms?Equivalence of categories ($c^*$ algebras <-> topological spaces)Category Theory compared with Meta-Grammars (or Hyper-Grammars) in Programming LanguagesCategory of sets equivalent to category of finite dimensional vector spaces?What is the opposite category of $operatorname{Top}$?What is the dual category of topological spaces?Axiom of Choice in Category TheoryConstruction of the equivalence of the finite ordinal category and the category of finite sets













12












$begingroup$


This question is inspired by this Mathoverflow question.



Ignoring size issues, there is a natural way to view a category as a first-order structure in a finite language. In light of this we can ask about the computational complexity of the first-order theory of a given category. Note that a lot of important structure is lost when passing to the first-order level; nevertheless, it still seems interesting to me.



Plenty of categories are easily checked to have very complicated first-order theories - for example, assuming the axiom of choice in the background we can identify the finite sets as those for which every self-injection is a surjection, and since Cartesian products and disjoint unions give the usual arithmetic we have that the first-order theory of the category Sets is not computable.



Incidentally, choice isn't needed here, but the non-choice argument is a bit more tedious.



My question is:




Is the first-order theory of the category Top of topological spaces (with morphisms being continuous maps) computable?




My suspicion is that the answer is no. An obvious way to prove this would be to show that the finite discrete spaces are first-order characterizable here, since then we could run the same argument as for Sets; however, I don't see how to do this. (In particular, the notion of "compact objcet" of a category is not obviously first-order expressible, so that characterization of finite discrete spaces doesn't seem to help.)





There are weaker versions of this question which could be asked:




  • We could focus on the theory of the categorial equivalence class of Top - that is, the set of sentences true in every category which is categorially equivalent to Top. In general categorial equivalence does not imply elementary equivalence (since a two-element category can be categorially equivalent to a one-element category), so this is a nontrivial weakening.


  • We could also restrict attention to a subcollection of categories - that is, pin down some non-first-order categorial property $P$ of Top, and then ask for a first-order theory which distinguishes the theory of Top (or the theory of the categorial equivalence class of Top) within this class. For example, Lawvere's ETCS characterizes Sets up to categorial equivalence amongst the locally small complete categories.



I am also interested in comments along these lines; however, my main question is specifically about the first-order theory of Top.



EDIT: David Roberts brought to my attention Schlomiuk's paper An elementary theory of the category of topological spaces; among other things, this gives a computable (indeed, finite) theory characterizing Top up to categorial equivalence amongst complete categories, as well as in a more subtle sense (any such category is equivalent to Top$^mathfrak{S}$ for some model $mathfrak{S}$ of ETCS).










share|cite|improve this question











$endgroup$

















    12












    $begingroup$


    This question is inspired by this Mathoverflow question.



    Ignoring size issues, there is a natural way to view a category as a first-order structure in a finite language. In light of this we can ask about the computational complexity of the first-order theory of a given category. Note that a lot of important structure is lost when passing to the first-order level; nevertheless, it still seems interesting to me.



    Plenty of categories are easily checked to have very complicated first-order theories - for example, assuming the axiom of choice in the background we can identify the finite sets as those for which every self-injection is a surjection, and since Cartesian products and disjoint unions give the usual arithmetic we have that the first-order theory of the category Sets is not computable.



    Incidentally, choice isn't needed here, but the non-choice argument is a bit more tedious.



    My question is:




    Is the first-order theory of the category Top of topological spaces (with morphisms being continuous maps) computable?




    My suspicion is that the answer is no. An obvious way to prove this would be to show that the finite discrete spaces are first-order characterizable here, since then we could run the same argument as for Sets; however, I don't see how to do this. (In particular, the notion of "compact objcet" of a category is not obviously first-order expressible, so that characterization of finite discrete spaces doesn't seem to help.)





    There are weaker versions of this question which could be asked:




    • We could focus on the theory of the categorial equivalence class of Top - that is, the set of sentences true in every category which is categorially equivalent to Top. In general categorial equivalence does not imply elementary equivalence (since a two-element category can be categorially equivalent to a one-element category), so this is a nontrivial weakening.


    • We could also restrict attention to a subcollection of categories - that is, pin down some non-first-order categorial property $P$ of Top, and then ask for a first-order theory which distinguishes the theory of Top (or the theory of the categorial equivalence class of Top) within this class. For example, Lawvere's ETCS characterizes Sets up to categorial equivalence amongst the locally small complete categories.



    I am also interested in comments along these lines; however, my main question is specifically about the first-order theory of Top.



    EDIT: David Roberts brought to my attention Schlomiuk's paper An elementary theory of the category of topological spaces; among other things, this gives a computable (indeed, finite) theory characterizing Top up to categorial equivalence amongst complete categories, as well as in a more subtle sense (any such category is equivalent to Top$^mathfrak{S}$ for some model $mathfrak{S}$ of ETCS).










    share|cite|improve this question











    $endgroup$















      12












      12








      12


      4



      $begingroup$


      This question is inspired by this Mathoverflow question.



      Ignoring size issues, there is a natural way to view a category as a first-order structure in a finite language. In light of this we can ask about the computational complexity of the first-order theory of a given category. Note that a lot of important structure is lost when passing to the first-order level; nevertheless, it still seems interesting to me.



      Plenty of categories are easily checked to have very complicated first-order theories - for example, assuming the axiom of choice in the background we can identify the finite sets as those for which every self-injection is a surjection, and since Cartesian products and disjoint unions give the usual arithmetic we have that the first-order theory of the category Sets is not computable.



      Incidentally, choice isn't needed here, but the non-choice argument is a bit more tedious.



      My question is:




      Is the first-order theory of the category Top of topological spaces (with morphisms being continuous maps) computable?




      My suspicion is that the answer is no. An obvious way to prove this would be to show that the finite discrete spaces are first-order characterizable here, since then we could run the same argument as for Sets; however, I don't see how to do this. (In particular, the notion of "compact objcet" of a category is not obviously first-order expressible, so that characterization of finite discrete spaces doesn't seem to help.)





      There are weaker versions of this question which could be asked:




      • We could focus on the theory of the categorial equivalence class of Top - that is, the set of sentences true in every category which is categorially equivalent to Top. In general categorial equivalence does not imply elementary equivalence (since a two-element category can be categorially equivalent to a one-element category), so this is a nontrivial weakening.


      • We could also restrict attention to a subcollection of categories - that is, pin down some non-first-order categorial property $P$ of Top, and then ask for a first-order theory which distinguishes the theory of Top (or the theory of the categorial equivalence class of Top) within this class. For example, Lawvere's ETCS characterizes Sets up to categorial equivalence amongst the locally small complete categories.



      I am also interested in comments along these lines; however, my main question is specifically about the first-order theory of Top.



      EDIT: David Roberts brought to my attention Schlomiuk's paper An elementary theory of the category of topological spaces; among other things, this gives a computable (indeed, finite) theory characterizing Top up to categorial equivalence amongst complete categories, as well as in a more subtle sense (any such category is equivalent to Top$^mathfrak{S}$ for some model $mathfrak{S}$ of ETCS).










      share|cite|improve this question











      $endgroup$




      This question is inspired by this Mathoverflow question.



      Ignoring size issues, there is a natural way to view a category as a first-order structure in a finite language. In light of this we can ask about the computational complexity of the first-order theory of a given category. Note that a lot of important structure is lost when passing to the first-order level; nevertheless, it still seems interesting to me.



      Plenty of categories are easily checked to have very complicated first-order theories - for example, assuming the axiom of choice in the background we can identify the finite sets as those for which every self-injection is a surjection, and since Cartesian products and disjoint unions give the usual arithmetic we have that the first-order theory of the category Sets is not computable.



      Incidentally, choice isn't needed here, but the non-choice argument is a bit more tedious.



      My question is:




      Is the first-order theory of the category Top of topological spaces (with morphisms being continuous maps) computable?




      My suspicion is that the answer is no. An obvious way to prove this would be to show that the finite discrete spaces are first-order characterizable here, since then we could run the same argument as for Sets; however, I don't see how to do this. (In particular, the notion of "compact objcet" of a category is not obviously first-order expressible, so that characterization of finite discrete spaces doesn't seem to help.)





      There are weaker versions of this question which could be asked:




      • We could focus on the theory of the categorial equivalence class of Top - that is, the set of sentences true in every category which is categorially equivalent to Top. In general categorial equivalence does not imply elementary equivalence (since a two-element category can be categorially equivalent to a one-element category), so this is a nontrivial weakening.


      • We could also restrict attention to a subcollection of categories - that is, pin down some non-first-order categorial property $P$ of Top, and then ask for a first-order theory which distinguishes the theory of Top (or the theory of the categorial equivalence class of Top) within this class. For example, Lawvere's ETCS characterizes Sets up to categorial equivalence amongst the locally small complete categories.



      I am also interested in comments along these lines; however, my main question is specifically about the first-order theory of Top.



      EDIT: David Roberts brought to my attention Schlomiuk's paper An elementary theory of the category of topological spaces; among other things, this gives a computable (indeed, finite) theory characterizing Top up to categorial equivalence amongst complete categories, as well as in a more subtle sense (any such category is equivalent to Top$^mathfrak{S}$ for some model $mathfrak{S}$ of ETCS).







      general-topology logic category-theory computability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 12 hours ago







      Noah Schweber

















      asked 13 hours ago









      Noah SchweberNoah Schweber

      126k10150289




      126k10150289






















          2 Answers
          2






          active

          oldest

          votes


















          11












          $begingroup$

          Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Yto X$ admits a section $g:Xto Y$ (i.e. $gcirc f=mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.



          If $X$ is discrete, then for any epimorphism $Yto X$ there is a set-theoretical section by AC, which is automatically continuous.



          Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Yto X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.



          The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.



          It's not clear to me if this proof can be modified so as to avoid the use of AC.



          Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1to 2$ any map. Then $X$ is discrete iff for any map $f:1to X$ there is a map $g:Xto 2$ such that $gcirc f=1$, and for any $f':1to X,f'neq f$, we have $gcirc f'neq f$.



          This expresses that for any point $xin X$, there is a map $Xto 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies ${x}$ is open, so $X$ is discrete.



          Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            @NoahSchweber Added a different, AC-free proof :)
            $endgroup$
            – Wojowu
            12 hours ago










          • $begingroup$
            Beautiful - accepted.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            Per your last line, see Eric's answer, which is what I had in mind for Sets.
            $endgroup$
            – Noah Schweber
            11 hours ago



















          6












          $begingroup$

          Here is a method of defining natural numbers in $mathbf{Set}$ (without Choice) which generalizes immediately to $mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1to N$ is a morphism, and $s:Nto N$ is a morphism such that they satisfy the second-order Peano axioms, namely:





          • $s$ is injective (i.e., monic).


          • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $scirc f$ for any $f:1to N$).

          • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:Ato N$ is any monic morphism such that $0$ factors through $i$ and $scirc i$ factors through $i$, then $i$ is an isomorphism.


          For $mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:Nto N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:Ato N$, and then the third condition implies $i$ is an isomorphism).



          (As an aside, while these axioms work in $mathbf{Set}$ or $mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)



          We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1to N$. We can define morphisms $+:Ntimes Nto N$ and $cdot:Ntimes Nto N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
            $endgroup$
            – Noah Schweber
            11 hours ago











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3133963%2fis-the-theory-of-the-category-of-topological-spaces-computable%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          11












          $begingroup$

          Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Yto X$ admits a section $g:Xto Y$ (i.e. $gcirc f=mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.



          If $X$ is discrete, then for any epimorphism $Yto X$ there is a set-theoretical section by AC, which is automatically continuous.



          Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Yto X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.



          The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.



          It's not clear to me if this proof can be modified so as to avoid the use of AC.



          Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1to 2$ any map. Then $X$ is discrete iff for any map $f:1to X$ there is a map $g:Xto 2$ such that $gcirc f=1$, and for any $f':1to X,f'neq f$, we have $gcirc f'neq f$.



          This expresses that for any point $xin X$, there is a map $Xto 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies ${x}$ is open, so $X$ is discrete.



          Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            @NoahSchweber Added a different, AC-free proof :)
            $endgroup$
            – Wojowu
            12 hours ago










          • $begingroup$
            Beautiful - accepted.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            Per your last line, see Eric's answer, which is what I had in mind for Sets.
            $endgroup$
            – Noah Schweber
            11 hours ago
















          11












          $begingroup$

          Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Yto X$ admits a section $g:Xto Y$ (i.e. $gcirc f=mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.



          If $X$ is discrete, then for any epimorphism $Yto X$ there is a set-theoretical section by AC, which is automatically continuous.



          Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Yto X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.



          The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.



          It's not clear to me if this proof can be modified so as to avoid the use of AC.



          Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1to 2$ any map. Then $X$ is discrete iff for any map $f:1to X$ there is a map $g:Xto 2$ such that $gcirc f=1$, and for any $f':1to X,f'neq f$, we have $gcirc f'neq f$.



          This expresses that for any point $xin X$, there is a map $Xto 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies ${x}$ is open, so $X$ is discrete.



          Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            @NoahSchweber Added a different, AC-free proof :)
            $endgroup$
            – Wojowu
            12 hours ago










          • $begingroup$
            Beautiful - accepted.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            Per your last line, see Eric's answer, which is what I had in mind for Sets.
            $endgroup$
            – Noah Schweber
            11 hours ago














          11












          11








          11





          $begingroup$

          Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Yto X$ admits a section $g:Xto Y$ (i.e. $gcirc f=mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.



          If $X$ is discrete, then for any epimorphism $Yto X$ there is a set-theoretical section by AC, which is automatically continuous.



          Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Yto X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.



          The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.



          It's not clear to me if this proof can be modified so as to avoid the use of AC.



          Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1to 2$ any map. Then $X$ is discrete iff for any map $f:1to X$ there is a map $g:Xto 2$ such that $gcirc f=1$, and for any $f':1to X,f'neq f$, we have $gcirc f'neq f$.



          This expresses that for any point $xin X$, there is a map $Xto 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies ${x}$ is open, so $X$ is discrete.



          Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.






          share|cite|improve this answer











          $endgroup$



          Let me assume choice. It is possible to distinguish discrete spaces from other spaces: those are precisely the spaces $X$ such that for any epimorphism $f:Yto X$ admits a section $g:Xto Y$ (i.e. $gcirc f=mathrm{id}_Y$). To see this, recall first that epimorphisms in Top are just the surjective continuous maps.



          If $X$ is discrete, then for any epimorphism $Yto X$ there is a set-theoretical section by AC, which is automatically continuous.



          Conversely, suppose every epimorphism into $X$ splits. Let $Y$ have the same underlying space as $X$, but discrete topology. Then the identity on points is a continuous map from $Yto X$, and since it's an epimorphism, it admits a section, which shows that $X$ is discrete too.



          The subcategory of discrete topological spaces is equivalent to the category of sets, and you already know it's uncomputable. It follows the first-order theory of Top is undecidable.



          It's not clear to me if this proof can be modified so as to avoid the use of AC.



          Edit: a proof without the use of AC: we will again distinguish the subcategory of discrete spaces. Let $1$ be a one-point space (a terminal object) and $2$ a two-point discrete space (a space with exactly two maps from $1$ to it, and with $4$ maps from it to itself) and $i:1to 2$ any map. Then $X$ is discrete iff for any map $f:1to X$ there is a map $g:Xto 2$ such that $gcirc f=1$, and for any $f':1to X,f'neq f$, we have $gcirc f'neq f$.



          This expresses that for any point $xin X$, there is a map $Xto 2$ which takes $x$ to one point, and all other points to a different one. Since $2$ was discrete, this implies ${x}$ is open, so $X$ is discrete.



          Since you say you know Set has uncomputable theory even without AC, we deduce the same for Top.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 12 hours ago

























          answered 12 hours ago









          WojowuWojowu

          18.6k23172




          18.6k23172












          • $begingroup$
            Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            @NoahSchweber Added a different, AC-free proof :)
            $endgroup$
            – Wojowu
            12 hours ago










          • $begingroup$
            Beautiful - accepted.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            Per your last line, see Eric's answer, which is what I had in mind for Sets.
            $endgroup$
            – Noah Schweber
            11 hours ago


















          • $begingroup$
            Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            @NoahSchweber Added a different, AC-free proof :)
            $endgroup$
            – Wojowu
            12 hours ago










          • $begingroup$
            Beautiful - accepted.
            $endgroup$
            – Noah Schweber
            12 hours ago










          • $begingroup$
            Per your last line, see Eric's answer, which is what I had in mind for Sets.
            $endgroup$
            – Noah Schweber
            11 hours ago
















          $begingroup$
          Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
          $endgroup$
          – Noah Schweber
          12 hours ago




          $begingroup$
          Neat, I didn't think of that! I'll hold off on accepting for a bit to see if a choice-free answer can be whipped up.
          $endgroup$
          – Noah Schweber
          12 hours ago












          $begingroup$
          @NoahSchweber Added a different, AC-free proof :)
          $endgroup$
          – Wojowu
          12 hours ago




          $begingroup$
          @NoahSchweber Added a different, AC-free proof :)
          $endgroup$
          – Wojowu
          12 hours ago












          $begingroup$
          Beautiful - accepted.
          $endgroup$
          – Noah Schweber
          12 hours ago




          $begingroup$
          Beautiful - accepted.
          $endgroup$
          – Noah Schweber
          12 hours ago












          $begingroup$
          Per your last line, see Eric's answer, which is what I had in mind for Sets.
          $endgroup$
          – Noah Schweber
          11 hours ago




          $begingroup$
          Per your last line, see Eric's answer, which is what I had in mind for Sets.
          $endgroup$
          – Noah Schweber
          11 hours ago











          6












          $begingroup$

          Here is a method of defining natural numbers in $mathbf{Set}$ (without Choice) which generalizes immediately to $mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1to N$ is a morphism, and $s:Nto N$ is a morphism such that they satisfy the second-order Peano axioms, namely:





          • $s$ is injective (i.e., monic).


          • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $scirc f$ for any $f:1to N$).

          • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:Ato N$ is any monic morphism such that $0$ factors through $i$ and $scirc i$ factors through $i$, then $i$ is an isomorphism.


          For $mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:Nto N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:Ato N$, and then the third condition implies $i$ is an isomorphism).



          (As an aside, while these axioms work in $mathbf{Set}$ or $mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)



          We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1to N$. We can define morphisms $+:Ntimes Nto N$ and $cdot:Ntimes Nto N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
            $endgroup$
            – Noah Schweber
            11 hours ago
















          6












          $begingroup$

          Here is a method of defining natural numbers in $mathbf{Set}$ (without Choice) which generalizes immediately to $mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1to N$ is a morphism, and $s:Nto N$ is a morphism such that they satisfy the second-order Peano axioms, namely:





          • $s$ is injective (i.e., monic).


          • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $scirc f$ for any $f:1to N$).

          • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:Ato N$ is any monic morphism such that $0$ factors through $i$ and $scirc i$ factors through $i$, then $i$ is an isomorphism.


          For $mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:Nto N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:Ato N$, and then the third condition implies $i$ is an isomorphism).



          (As an aside, while these axioms work in $mathbf{Set}$ or $mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)



          We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1to N$. We can define morphisms $+:Ntimes Nto N$ and $cdot:Ntimes Nto N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
            $endgroup$
            – Noah Schweber
            11 hours ago














          6












          6








          6





          $begingroup$

          Here is a method of defining natural numbers in $mathbf{Set}$ (without Choice) which generalizes immediately to $mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1to N$ is a morphism, and $s:Nto N$ is a morphism such that they satisfy the second-order Peano axioms, namely:





          • $s$ is injective (i.e., monic).


          • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $scirc f$ for any $f:1to N$).

          • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:Ato N$ is any monic morphism such that $0$ factors through $i$ and $scirc i$ factors through $i$, then $i$ is an isomorphism.


          For $mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:Nto N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:Ato N$, and then the third condition implies $i$ is an isomorphism).



          (As an aside, while these axioms work in $mathbf{Set}$ or $mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)



          We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1to N$. We can define morphisms $+:Ntimes Nto N$ and $cdot:Ntimes Nto N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.






          share|cite|improve this answer











          $endgroup$



          Here is a method of defining natural numbers in $mathbf{Set}$ (without Choice) which generalizes immediately to $mathbf{Top}$. Namely, we can just express the second-order Peano axioms directly in our categorical language. Let us fix a terminal object $1$. Then we can define the natural numbers as a triple $(N,s,0)$ where $N$ is an object, $0:1to N$ is a morphism, and $s:Nto N$ is a morphism such that they satisfy the second-order Peano axioms, namely:





          • $s$ is injective (i.e., monic).


          • $0$ is not in the image of $s$ (that is, $0$ cannot be factored as a composition $scirc f$ for any $f:1to N$).

          • No proper subobject of $N$ contains $0$ and is closed under $s$. That is, if $i:Ato N$ is any monic morphism such that $0$ factors through $i$ and $scirc i$ factors through $i$, then $i$ is an isomorphism.


          For $mathbf{Set}$, these properties pick out the usual structure of natural numbers (up to isomorphism), and for $mathbf{Top}$, they pick out the natural numbers with the discrete topology (proof: given any space $N$ with element $0$ and map $s:Nto N$ satisfying the first two conditions, the subset generated by $0$ under $s$ can be given the discrete topology to get a space $A$ and a continuous injection $i:Ato N$, and then the third condition implies $i$ is an isomorphism).



          (As an aside, while these axioms work in $mathbf{Set}$ or $mathbf{Top}$, there are other axioms for natural numbers besides the Peano axioms which are more natural from a categorical perspective, such as the notion of a natural numbers object.)



          We can then fix any such $(N,s,0)$ and define a natural number as a morphism $1to N$. We can define morphisms $+:Ntimes Nto N$ and $cdot:Ntimes Nto N$ as the unique morphisms satisfying the usual recursive definitions, and these give binary operations on our natural numbers, so we can interpret arithmetic in our category.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 11 hours ago

























          answered 11 hours ago









          Eric WofseyEric Wofsey

          188k14216345




          188k14216345












          • $begingroup$
            This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
            $endgroup$
            – Noah Schweber
            11 hours ago


















          • $begingroup$
            This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
            $endgroup$
            – Noah Schweber
            11 hours ago
















          $begingroup$
          This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
          $endgroup$
          – Noah Schweber
          11 hours ago




          $begingroup$
          This was the argument I had in mind for the choice-free version w/r/t Sets - stupid though it seems, I didn't notice that it worked for Top as well!
          $endgroup$
          – Noah Schweber
          11 hours ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3133963%2fis-the-theory-of-the-category-of-topological-spaces-computable%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          El tren de la libertad Índice Antecedentes "Porque yo decido" Desarrollo de la...

          Castillo d'Acher Características Menú de navegación

          Connecting two nodes from the same mother node horizontallyTikZ: What EXACTLY does the the |- notation for...