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Here's my reasoning:

Let X be a connected set. Assume, for contradiction, that X is not perfect. Then there exists a point p in X and a neighborhood N of p such that the intersection of N and X is equal to {p}. This implies that N is both open and closed in X, making it a clopen subset of X. However, the existence of a non-trivial clopen subset contradicts the assumption that X is connected. Thus, we conclude that X must be perfect.

Is my reasoning correct, or am I missing something?

MiFedorov
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    You're assuming every singleton is closed, which is not true in every topological space. However your proof seems valid for $T_1$ spaces. – Steven Clontz Dec 25 '24 at 01:34
  • You are missing a definition of "perfect set." The usual definition of perfect set is a set which is equal to its defived set, in other words, a closed set with no isolated points. The open interval $(0,1)$ is a connected subset of the real line but it's not a perfect set since it's not closed. If you are using the term "perfect set" in some other sense, you should define what you mean by it. – user14111 Dec 25 '24 at 01:45

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Your reasoning is not correct. There's no reason to believe that $N\cap X=\{p\}$ implies that $N$ is closed. Besides that, you're trying to actually prove that a connected space has no isolated points, and even this is not true, for instance the topological space consisting of just one point is connected but has isolated points. Outside of that, you can fix your argument to prove that if a space with at least two elements where points are closed and is connected, then it has no isolated points, by concluding that $\{p\}$ is clopen.

That being said, for a set to be perfect, as noted in the comments, you need the set to be closed. Since every space is closed with respect to itself you get that for free, but in general, a set might be connected but not perfect, even if points are closed, for instance, $(0,1)$ is a connected subset of $[0,1]$ but it's not perfect as it's not closed.

All in all, you can tweak your arguments so that you prove that in every connected space where points are closed you have no isolated points (which in turn makes it perfect with respect to itself), but not that a subset of a topological space is perfect just by being connected, even in the case where points are closed

Bruno Andrades
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  • Could you please elaborate how N∩X={p} doesn't imply that N is closed with respect to X? Suppose x (in X) is a limit point of N. Then any neighborhood of x would contain a point of N - namely p. But then any neigbourhood of p would contain x. Since I've found a neighborhood of p that doesn't contain any element other than p, it doesn't contain x, so x is not a limit point of N. So N has no limit points and is thus closed. But yeah, I see now how my proof was only concerning the fact that in a connected space with at least 2 points, there are no isolated points rather than that it's perfect. – MiFedorov Dec 26 '24 at 08:04
  • @MiFedorov I think I know what you're trying to do, but you got it backwards. I think what you want to show is that $N$ contains its limit points to show it's closed. The problem with that is that the way you phrased it you seem to be trying to prove that $N$ doesn't have accumulation points (which is technically not the same as a limit point) which doesn't have anything to do with being closed or open – Bruno Andrades Dec 26 '24 at 11:30
  • I'm not sure... This is my first intro to real analysis and I'm using Rudin's definition of a limit point (he hasn't defined the concept of an accumulation point). Also, from this response https://math.stackexchange.com/a/1559027/1318759 it seems like I've actually shown that N doesn't have limit points (rather than that it doesn't have accumulation points). That there are no limit points (in the context of the space X rather than the entire universe) also follows from N being finite (Rudin has a proof regarding that finite sets cannot having limit points) – MiFedorov Dec 26 '24 at 21:18