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Upon reviewing Rudin's basic topology section, I was asking myself:"Is every interior point limit point?" Because at first it seems like an interior point $p$ of a set $E$ always contain something in $E$ that is not $p$ itself.

But then I think if the discrete finite set $E$, for which each point is an interior point but not limit point because it is a finite set.

Then I thought what if $E$ is connected, not discrete -- does this mean $E$ is not finite? Then does this mean every interior point is actually a limit point of $E$?

Comments are much appreciated! Intuition tells me yes, but I don't have enough techniques yet to prove. Thanks!

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    $\begingroup$ It is not true that every point in a discrete finite set is an interior point - in fact, such sets (usually) have no interior points! $\endgroup$ – Noah Schweber Nov 12 '15 at 8:54
  • $\begingroup$ See Sierpiński space for a connected finite set (discrete topology with at least two points is never connected). $\endgroup$ – Jean-Claude Arbaut Nov 12 '15 at 8:55
  • $\begingroup$ To talk about interior points of $E$ you have to specify the topological space you consider $E$ to be a subset of. If this is $E$ itself, every point is an interior point! $\endgroup$ – Christoph Nov 12 '15 at 8:56
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In any metric spaces $(X,d)$ (which is what Rudin dealt with mostly in his PoMA), every interior points of a connected set is a limit point. To see why, observe that for $x\in C^o$, where $C$ is a connected subset of $X$, there exist $r>0$ such that $x\in B(x,r)\subset C$.

For any $D\in \Bbb R$ such that $0<D<r$, there exist $y_D\in B(x,r)$ such that $d(x,y_D)=D$ or else $C$ can be "disconnected" by $$ C = \{y\in C|d(x,y)<D\}\cup \{y\in C|d(x,y)>D\}\ $$

By letting $D\to 0$ we have $y_D\to x$ so $x$ is indeed a limit point of $C$.

Edited: As Mr. Fischer has pointed out, the connected set need to consist of more than 1 element in order for the argument to work. If $x\in X$ is an isolated point then the singleton $\{x\}$ is a connected set which is, obviously, not a limit point.

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    $\begingroup$ If $x$ is an isolated point of $X$, then $\{x\}$ is a connected open set without limit points. $\endgroup$ – Daniel Fischer Nov 12 '15 at 13:27
  • $\begingroup$ @DanielFischer You are absolutely right, I totally neglected the case where the open set is a singleton. I will edit accordingly. $\endgroup$ – BigbearZzz Nov 12 '15 at 13:54
  • $\begingroup$ @DanielFischer If $x$ is an isolated point of $X$, is $\{x\}$ the only kind of finite connected set in $X$? $\endgroup$ – nekodesu Nov 12 '15 at 20:55
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    $\begingroup$ @ShiyueLi In metric spaces (more generally, in $T_1$ spaces), the only finite connected sets are singletons and the empty set [caution, however: there are people who don't consider the empty set connected]. That $x$ is an isolated point makes $\{x\}$ open, so it has an interior point. If $y$ is a non-isolated point, $\{y\}$ is connected, but not open, and $\{y\}$ has no interior points. And that means that vacuously every interior point of $\{y\}$ is a limit point of $\{y\}$. $\endgroup$ – Daniel Fischer Nov 12 '15 at 21:09
  • $\begingroup$ @DanielFischer Thank you for your prompt response! I'm not sure I understand the part about $x$ being isolated point. If $x$ is an isolated point, that means $x \in \{x\}$ and $x$ is not a limit point of $\{x\}$. How does this make $\{x\}$ open? Because I thought $x$ has no neighborhood that $N_r$ for $r > 0$ such that $N_r \subset \{x\}$, which means $x$ is not an interior point so $ \{x\}$ is not open...Really appreciate it if you could correct my reasoning. Also, if $y$ is non-isolated point, why is $ \{y\}$ connected but not open. Since I thought $ \{y\}$ is always not open. $\endgroup$ – nekodesu Nov 12 '15 at 21:17
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Yes, this holds in $T_1$ spaces, i.e. topological spaces where all singletons are closed. This is certainly true in metric spaces, as for $y \neq x$ we have that $B(y, d(x,y) ) \cap \{x\}$ is empty, showing that all $y \in X \setminus \{x\}$ are interior points.

Firstly $T_1$ implies that all finite sets are closed (as finite unions of closed singletons). And so every finite $F$ is discrete in itself, so disconnected if it has 2 or more points. Hence connected sets are infinite.

Also if $C$ is connected and has more than one point, and if $x$ is in the interior $C^{\circ}$ (it could also be empty, of course, like a circle in the plane) then $x$ is a limit point of $C$. If not, $\{x\}$ would be open in $C$, and $C$ would not be connected (as $\{x\}$ and $C \setminus \{x\}$ would separate it).

Of course, if $X$ has an isolated point, $\{x\}$ would be connected and $x$ would be in its interior and no limit point, so we do need more than one point in $C$.

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