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I refer to the answer in the following link as the source of my question.

In a metric space X, if A is connected, is its interior connected?

While I understand the counterexample given, I actually thought that the interior of a connected set is connected. My proof is as follows:

Since $A$ is a connected set, suppose we take any two-valued function $f$ on $A$. Then, for any $a \in A$, $f(a)=c$ for some constant $c$. Since $A^{\circ} \subset A$, the restriction of $f$ on $A^{\circ} $ is also constant, that is, $f|_{A^{\circ}}=c$ and hence the interior of a connected set is connected.

Does the two-valued function argument only work on $\mathbb{R}$ or am I missing something fundamental? Thank you.

Arjun-Gurung
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    May be there are two-valued continuous functions defined on $A^\circ$ that cannot be gotten by restricting a two-valued continuous function on all of $A$? – Jyrki Lahtonen Nov 08 '23 at 20:58
  • I agree with @JyrkiLahtonen – Djalal Ounadjela Nov 08 '23 at 21:01
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    If $A\subseteq\mathbb{R}^n$ is convex, then $A^\circ$ is convex. So its true for convex subsets of $\mathbb{R}^n$. In particular, its true for $\mathbb{R}$, since $A\subseteq \mathbb{R}$ is convex iff its connected. – Jakobian Nov 08 '23 at 21:08
  • @JyrkiLahtonen Should it not be the other way around since we are assuming $A$ is connected and not $A^{\circ}$? I mean should it not be that I cannot restrict $f$ on $A^{\circ}$ because $f$ might lose its continuity? – Arjun-Gurung Nov 08 '23 at 21:26
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    Generally if you've written a proof of a false statement, it's good practice to try and "run" your proof on a counterexample showing the statement is false. In this case, hopefully you agree that the interior of the two discs is not connected? So you can define a function which is $0$ on the interior of the left disc but $1$ on the interior of the right disc. You're claiming you've proved this isn't possible. Can you see why your proof doesn't actually show that's impossible? – Izaak van Dongen Nov 08 '23 at 21:30
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    Nope. You want to prove that $A^\circ$ is connected. To apply the theorem with two-valued functions you need to consider all the two-valued continuous functions defined on $A^\circ$. In your argument you only consider those two-valued functions that are gotten by restricting a two-valued function continuous on all of $A$. – Jyrki Lahtonen Nov 08 '23 at 21:41
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    And the process of restriction always preserves continuity (assuming the usual subspace topology). The trouble begins when you want to extend a function from a smaller domain to a bigger one while preserving continuity. – Jyrki Lahtonen Nov 08 '23 at 21:44
  • @IzaakvanDongen I think I see your point. If I restrict the function onto the interior of the left disc (or the right disc respectively), then I will get a constant function but the interior of the two tangent discs is not connected. So in sum of all the comments here, I need to start with the connectedness of $A^{\circ}$ and then work upwards to the connectedness of $A$ to derive a contradiction. – Arjun-Gurung Nov 09 '23 at 06:17
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    Well, if you wanted to prove the interior was always connected, the argument would have to go "suppose the interior is disconnected. .... therefore $A$ is disconnected, which is absurd". So if you're arguing via functions, the argument would have to go "let $f$ be a two-valued continuous surjection on the interior. ... Therefore $g$ is a two-valued continuous surjection on $A$, which is absurd". Your argument is "let $f$ be a two-valued function on $A$. ... Therefore $g$ is a two-valued constant function on the interior." This just proves some functions on the interior are constant - not all! – Izaak van Dongen Nov 09 '23 at 11:51

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If that argument was correct, it prove not only $\mathring A$ is connected but, in fact, that every subset of $A$ is connected.

The problem lies in the implicit assumption that every continuous function $f\colon\mathring A\longrightarrow\{0,1\}$ is the restriction to $\mathring A$ of some continuous function from $A$ into $\{0,1\}$. This is false.

José Carlos Santos
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  • Could you please clarify on the second part of your answer in light of the this link? https://math.stackexchange.com/questions/1826827/topology-show-restriction-of-continuous-function-is-continuous-and-restriction . I am wondering if it is a matter of continuity not holding on the restricted function or if it is that the set in the metric space going from closed to open. – Arjun-Gurung Nov 08 '23 at 21:22
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    I don't see what my answer has to do with the content of that link. That link has to do with the connection between $f$ and $f|A$. But the problem with your approach is that you seem to think (wrongly) that every continuous function from $\mathring A$ into ${0,1}$ is of the form $f|{\mathring A}$, for some continuous function $f\colon A\longrightarrow{0,1}$. – José Carlos Santos Nov 08 '23 at 21:40
  • I understand now, thank you very much! – Arjun-Gurung Nov 09 '23 at 06:11