Mixing
Is it possible to put a topology on Turing-recognizable languages to express density among all the languages?
$begingroup$
In a Calculability and complexity course I had at univeristy, we proved that there exist languages that are not Turing-recognizable basiclly using Cantor's diagonal argument (the set of all languages is uncountable and the set of Turing-recognizable languages is countable). This immediately brought the analogy with $mathbb Q$ which is countable inside $mathbb R$ which is not. But we have a topology on $mathbb R$ (and thus on $mathbb Q$) which allows us to show that $mathbb Q$ is a dense subset of $mathbb R$ (also it's a metric space).
A question then popped in my head: is there any way to put a topology onto the set of languages that would lead to a smilar result, i.e. density of the Turing-recognizable lanuages among all the languages?
Note that I have no idea if this even makes sense: I'm not really into theoretical computing, and I have no idea if the notion of "closeness" between languages makes any sense. This is just a question I've hold for too long now and I had to ask if somebody already answered this or if this is just a pointless question.
general-topology computer-science computability turing-machines
asked Jan 12 at 0:01
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
In a Calculability and complexity course I had at univeristy, we proved that there exist languages that are not Turing-recognizable basiclly using Cantor's diagonal argument (the set of all languages is uncountable and the set of Turing-recognizable languages is countable). This immediately brought the analogy with $mathbb Q$ which is countable inside $mathbb R$ which is not. But we have a topology on $mathbb R$ (and thus on $mathbb Q$) which allows us to show that $mathbb Q$ is a dense subset of $mathbb R$ (also it's a metric space).
A question then popped in my head: is there any way to put a topology onto the set of languages that would lead to a smilar result, i.e. density of the Turing-recognizable lanuages among all the languages?
Note that I have no idea if this even makes sense: I'm not really into theoretical computing, and I have no idea if the notion of "closeness" between languages makes any sense. This is just a question I've hold for too long now and I had to ask if somebody already answered this or if this is just a pointless question.
general-topology computer-science computability turing-machines
asked Jan 12 at 0:01
BermudesBermudes
18713
$endgroup$
add a comment |
$begingroup$
In a Calculability and complexity course I had at univeristy, we proved that there exist languages that are not Turing-recognizable basiclly using Cantor's diagonal argument (the set of all languages is uncountable and the set of Turing-recognizable languages is countable). This immediately brought the analogy with $mathbb Q$ which is countable inside $mathbb R$ which is not. But we have a topology on $mathbb R$ (and thus on $mathbb Q$) which allows us to show that $mathbb Q$ is a dense subset of $mathbb R$ (also it's a metric space).
A question then popped in my head: is there any way to put a topology onto the set of languages that would lead to a smilar result, i.e. density of the Turing-recognizable lanuages among all the languages?
Note that I have no idea if this even makes sense: I'm not really into theoretical computing, and I have no idea if the notion of "closeness" between languages makes any sense. This is just a question I've hold for too long now and I had to ask if somebody already answered this or if this is just a pointless question.
general-topology computer-science computability turing-machines
asked Jan 12 at 0:01
BermudesBermudes
18713
$endgroup$
In a Calculability and complexity course I had at univeristy, we proved that there exist languages that are not Turing-recognizable basiclly using Cantor's diagonal argument (the set of all languages is uncountable and the set of Turing-recognizable languages is countable). This immediately brought the analogy with $mathbb Q$ which is countable inside $mathbb R$ which is not. But we have a topology on $mathbb R$ (and thus on $mathbb Q$) which allows us to show that $mathbb Q$ is a dense subset of $mathbb R$ (also it's a metric space).
A question then popped in my head: is there any way to put a topology onto the set of languages that would lead to a smilar result, i.e. density of the Turing-recognizable lanuages among all the languages?
Note that I have no idea if this even makes sense: I'm not really into theoretical computing, and I have no idea if the notion of "closeness" between languages makes any sense. This is just a question I've hold for too long now and I had to ask if somebody already answered this or if this is just a pointless question.
general-topology computer-science computability turing-machines
general-topology computer-science computability turing-machines
asked Jan 12 at 0:01
BermudesBermudes
18713
asked Jan 12 at 0:01
BermudesBermudes
18713
asked Jan 12 at 0:01
BermudesBermudes
18713
asked Jan 12 at 0:01
BermudesBermudes
18713
asked Jan 12 at 0:01
BermudesBermudes
18713
18713
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From a slightly different perspective: for theoretical purposes it's often better to work with computable and uncomputable sets rather than languages; indeed, if your language is over a finite alphabet $A$, then there's the classic natural correspondance between $A^*$ and $mathbb{N}$ given by 'base-$|A|$ representation', and it shouldn't be hard to convince yourself that a language $mathcal{L}subseteqmathcal{P}(A^*)$ is Turing recognizable iff the set $S_Lsubseteqmathbb{N}$ given by the correspondence is computable.
This lets us use a standard topology on subsets of $mathbb{N}$; for instance, we can use a metric like $d(A,B)=2^{-n}$, where $n$ is the smallest element of $ADelta B$. Under this topology, the computable sets are indeed 'dense' in the set of all subsets of $mathbb{N}$ — in fact, the finite sets (all of which are computable) are already dense. (This is exactly analagous to the dyadic rationals being dense in the set of reals, although the 'natural' mapping $Ssubseteqmathbb{N}mapsto sum_{sin S} 2^{-s}$ to map the subsets of the naturals into $[0,1]$ doesn't quite work since it's not 1-to-1.)
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
$endgroup$
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
add a comment |
$begingroup$
I don’t know if this can be done as you want, but some thoughts: a language over some fixed alphabet $Sigma$, is just a set of (finite length) strings, so a member of $mathscr{P}(Sigma^ast)$. $Sigma^ast$ is essentially a subset of a product $Sigma^omega$ so can be given a somewhat natural topology (if the finite set $Sigma$ gets the discrete topology; we get a subset of a Cantor set, topologically).
The power set of $Sigma^ast$ can be given a product topology again, using the characteristic function identification. It is well-known that this set indeed has a countable dense subset (Hewitt-Marzcewksi-Pondizcery theorem) so I think there is good hope we can find a Turing-recognisable family of languages that is dense in this topology as described. No time to look into further details as of yet.. Looks like a nice exercise for someone, though.
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070465%2fis-it-possible-to-put-a-topology-on-turing-recognizable-languages-to-express-den%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
$begingroup$
From a slightly different perspective: for theoretical purposes it's often better to work with computable and uncomputable sets rather than languages; indeed, if your language is over a finite alphabet $A$, then there's the classic natural correspondance between $A^*$ and $mathbb{N}$ given by 'base-$|A|$ representation', and it shouldn't be hard to convince yourself that a language $mathcal{L}subseteqmathcal{P}(A^*)$ is Turing recognizable iff the set $S_Lsubseteqmathbb{N}$ given by the correspondence is computable.
This lets us use a standard topology on subsets of $mathbb{N}$; for instance, we can use a metric like $d(A,B)=2^{-n}$, where $n$ is the smallest element of $ADelta B$. Under this topology, the computable sets are indeed 'dense' in the set of all subsets of $mathbb{N}$ — in fact, the finite sets (all of which are computable) are already dense. (This is exactly analagous to the dyadic rationals being dense in the set of reals, although the 'natural' mapping $Ssubseteqmathbb{N}mapsto sum_{sin S} 2^{-s}$ to map the subsets of the naturals into $[0,1]$ doesn't quite work since it's not 1-to-1.)
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
$endgroup$
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
add a comment |
$begingroup$
From a slightly different perspective: for theoretical purposes it's often better to work with computable and uncomputable sets rather than languages; indeed, if your language is over a finite alphabet $A$, then there's the classic natural correspondance between $A^*$ and $mathbb{N}$ given by 'base-$|A|$ representation', and it shouldn't be hard to convince yourself that a language $mathcal{L}subseteqmathcal{P}(A^*)$ is Turing recognizable iff the set $S_Lsubseteqmathbb{N}$ given by the correspondence is computable.
This lets us use a standard topology on subsets of $mathbb{N}$; for instance, we can use a metric like $d(A,B)=2^{-n}$, where $n$ is the smallest element of $ADelta B$. Under this topology, the computable sets are indeed 'dense' in the set of all subsets of $mathbb{N}$ — in fact, the finite sets (all of which are computable) are already dense. (This is exactly analagous to the dyadic rationals being dense in the set of reals, although the 'natural' mapping $Ssubseteqmathbb{N}mapsto sum_{sin S} 2^{-s}$ to map the subsets of the naturals into $[0,1]$ doesn't quite work since it's not 1-to-1.)
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
$endgroup$
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
add a comment |
$begingroup$
From a slightly different perspective: for theoretical purposes it's often better to work with computable and uncomputable sets rather than languages; indeed, if your language is over a finite alphabet $A$, then there's the classic natural correspondance between $A^*$ and $mathbb{N}$ given by 'base-$|A|$ representation', and it shouldn't be hard to convince yourself that a language $mathcal{L}subseteqmathcal{P}(A^*)$ is Turing recognizable iff the set $S_Lsubseteqmathbb{N}$ given by the correspondence is computable.
This lets us use a standard topology on subsets of $mathbb{N}$; for instance, we can use a metric like $d(A,B)=2^{-n}$, where $n$ is the smallest element of $ADelta B$. Under this topology, the computable sets are indeed 'dense' in the set of all subsets of $mathbb{N}$ — in fact, the finite sets (all of which are computable) are already dense. (This is exactly analagous to the dyadic rationals being dense in the set of reals, although the 'natural' mapping $Ssubseteqmathbb{N}mapsto sum_{sin S} 2^{-s}$ to map the subsets of the naturals into $[0,1]$ doesn't quite work since it's not 1-to-1.)
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
$endgroup$
From a slightly different perspective: for theoretical purposes it's often better to work with computable and uncomputable sets rather than languages; indeed, if your language is over a finite alphabet $A$, then there's the classic natural correspondance between $A^*$ and $mathbb{N}$ given by 'base-$|A|$ representation', and it shouldn't be hard to convince yourself that a language $mathcal{L}subseteqmathcal{P}(A^*)$ is Turing recognizable iff the set $S_Lsubseteqmathbb{N}$ given by the correspondence is computable.
This lets us use a standard topology on subsets of $mathbb{N}$; for instance, we can use a metric like $d(A,B)=2^{-n}$, where $n$ is the smallest element of $ADelta B$. Under this topology, the computable sets are indeed 'dense' in the set of all subsets of $mathbb{N}$ — in fact, the finite sets (all of which are computable) are already dense. (This is exactly analagous to the dyadic rationals being dense in the set of reals, although the 'natural' mapping $Ssubseteqmathbb{N}mapsto sum_{sin S} 2^{-s}$ to map the subsets of the naturals into $[0,1]$ doesn't quite work since it's not 1-to-1.)
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
answered Jan 12 at 2:23
Steven StadnickiSteven Stadnicki
41.1k868122
41.1k868122
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
add a comment |
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
$begingroup$
Oh that is clever. However, I guess the topology would change depending on the mapping $A to [1, |A|] cap mathbb N$ though this won't change the result of density since the finite sets are dense. I also guess that changing this initial mapping will lead to a homeomorphic topology.
$endgroup$
– Bermudes
Jan 12 at 10:43
add a comment |
$begingroup$
I don’t know if this can be done as you want, but some thoughts: a language over some fixed alphabet $Sigma$, is just a set of (finite length) strings, so a member of $mathscr{P}(Sigma^ast)$. $Sigma^ast$ is essentially a subset of a product $Sigma^omega$ so can be given a somewhat natural topology (if the finite set $Sigma$ gets the discrete topology; we get a subset of a Cantor set, topologically).
The power set of $Sigma^ast$ can be given a product topology again, using the characteristic function identification. It is well-known that this set indeed has a countable dense subset (Hewitt-Marzcewksi-Pondizcery theorem) so I think there is good hope we can find a Turing-recognisable family of languages that is dense in this topology as described. No time to look into further details as of yet.. Looks like a nice exercise for someone, though.
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
add a comment |
$begingroup$
I don’t know if this can be done as you want, but some thoughts: a language over some fixed alphabet $Sigma$, is just a set of (finite length) strings, so a member of $mathscr{P}(Sigma^ast)$. $Sigma^ast$ is essentially a subset of a product $Sigma^omega$ so can be given a somewhat natural topology (if the finite set $Sigma$ gets the discrete topology; we get a subset of a Cantor set, topologically).
The power set of $Sigma^ast$ can be given a product topology again, using the characteristic function identification. It is well-known that this set indeed has a countable dense subset (Hewitt-Marzcewksi-Pondizcery theorem) so I think there is good hope we can find a Turing-recognisable family of languages that is dense in this topology as described. No time to look into further details as of yet.. Looks like a nice exercise for someone, though.
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
add a comment |
$begingroup$
I don’t know if this can be done as you want, but some thoughts: a language over some fixed alphabet $Sigma$, is just a set of (finite length) strings, so a member of $mathscr{P}(Sigma^ast)$. $Sigma^ast$ is essentially a subset of a product $Sigma^omega$ so can be given a somewhat natural topology (if the finite set $Sigma$ gets the discrete topology; we get a subset of a Cantor set, topologically).
The power set of $Sigma^ast$ can be given a product topology again, using the characteristic function identification. It is well-known that this set indeed has a countable dense subset (Hewitt-Marzcewksi-Pondizcery theorem) so I think there is good hope we can find a Turing-recognisable family of languages that is dense in this topology as described. No time to look into further details as of yet.. Looks like a nice exercise for someone, though.
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
$endgroup$
I don’t know if this can be done as you want, but some thoughts: a language over some fixed alphabet $Sigma$, is just a set of (finite length) strings, so a member of $mathscr{P}(Sigma^ast)$. $Sigma^ast$ is essentially a subset of a product $Sigma^omega$ so can be given a somewhat natural topology (if the finite set $Sigma$ gets the discrete topology; we get a subset of a Cantor set, topologically).
The power set of $Sigma^ast$ can be given a product topology again, using the characteristic function identification. It is well-known that this set indeed has a countable dense subset (Hewitt-Marzcewksi-Pondizcery theorem) so I think there is good hope we can find a Turing-recognisable family of languages that is dense in this topology as described. No time to look into further details as of yet.. Looks like a nice exercise for someone, though.
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
answered Jan 12 at 1:28
Henno BrandsmaHenno Brandsma
109k347114
109k347114
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
add a comment |
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
$begingroup$
There isn't a way I want to do it: this is just a question I had in my head for some reason, but I won't make any use of it besides feeling relieved and smarter at the same time. ;) It looks very nice though, especially since the Cantor set topology appears. I'll take a look at the topology we end up with on $mathcal P(Sigma^*)$ but I have never heard of the characteristic function identification.
$endgroup$
– Bermudes
Jan 12 at 10:50
1
1
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
$begingroup$
@Bermudes A subset $A$ of $X$ can be identified with the function $chi_A: X to {0,1}$ that sends $xin A$ to $1$ and all other $x$ to $0$.
$endgroup$
– Henno Brandsma
Jan 12 at 12:12
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070465%2fis-it-possible-to-put-a-topology-on-turing-recognizable-languages-to-express-den%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
