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While considering some examples like Topologist's sine curve and Another example of a connected but non path connected set It seems to me sometimes closure of path connected space may not be path connected because there are some limit point which is very near to an element but can not be connected by a finite path..and in spite of being the limit point and the element so close to each other ,one needs to trace infinite distance in order to go from the element to the limit point...

I don't know what I am thinking is absurd or not..But it seems to me this fact is responsible for being the closure of some path connected space is not path connected.is there any concept or theorems about the fact.please help.

Samiron Parui
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    That's a pretty reasonable intuition, at least in the specific case of the topologist's sine curve and related examples. – Arthur Nov 13 '17 at 15:53
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    Your intuition is pretty good, and it's usually formalized by making a compactness argument. That's usually how one says you can't cover that infinite distance because compactness often links up with boundedness. – Randall Nov 13 '17 at 16:05
  • Means are you saying If X is compact and path connected then the closure of X is also path connected?? – Samiron Parui Nov 13 '17 at 17:08

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A perfectly reasonable intuition, since this is the essential problem with the topologist's sin curve. In fact, a connected open subset of $\mathbb R^n$ is path-connected.

One can prove the last claim by considering $a \in U$ for some connected open subset, and then defining $H$ to be the set of points that can be joined in a path to $U$. First note, that since $U$ is open, there is some $B_x \subset U$ for any $x \in H$, and since the ball is convex, it is path-connected, so we can find a path for any $y \in B_x$ to $x$, and hence to $a$ by concatenation. Hence, $B_x \subset H$ in fact. Similarly, the complement of $H$ is open. Since $H$ is nonempty ($a \in H$), we can conclude that $U\setminus H$ is empty by the connected assumption, which implies that $U=H$, so it is path connected.

Andres Mejia
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  • Generalizing this, if a space is both connected and locally path connected, then it is connected. Locally path connected means that there is a basis of path connected open sets. $\mathbb{R}^n$ is locally path connected because the open balls are such a basis. – Michael Weiss Nov 13 '17 at 16:54
  • @MichaelWeiss nice point. – Andres Mejia Nov 13 '17 at 16:55
  • Is there any specific condition,under which the closure of a path connected set would always be path connected? – Samiron Parui Nov 13 '17 at 17:40
  • Not really that I know of... here is a related question https://math.stackexchange.com/questions/162867/when-is-the-closure-of-a-path-connected-set-also-path-connected – Andres Mejia Nov 13 '17 at 17:52