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Munkres problem 24.11: If $A$ is a connected subspace of $X$, does it follow that $\operatorname{Int}A$ and $\operatorname{Bd}A$ are connected? Does the converse hold?

I've answered these questions, but I'm wondering if the converse might hold if $X$ satisfies some reasonable conditions, like being normal and connected, or perhaps locally connected, and assuming that neither the interior nor the boundary is empty.

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2 Answers 2

up vote 9 down vote accepted

The converse fails to hold even in very pleasant topological spaces, like the reals. Of course, $\mathbb R$ is perfectly normal, connected, etc. However, even though $\mathbb Q \subseteq \mathbb R$ has both its boundary (namely, $\mathbb R$ itself) and its interior (the empty interior) connected, the set $\mathbb Q$ itself is disconnected.


Tweak to get nonempty interior: The OP desired a set $A$ with nonempty interior in the comments. For this, take the disconnected set $A = (-\infty, 0] \cup \mathbb Q_{\geqslant 0}$. However, its interior $\operatorname{Int} A = (-\infty, 0)$ and its boundary $\operatorname{Bd} A = [0, \infty)$ are both nonempty and connected.

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I hadn't thought about the option of the interior being null. I'll exclude it. –  dfeuer Dec 4 '11 at 5:16
    
Ah, that doesn't help anything, does it? You get the checkmark. –  dfeuer Dec 4 '11 at 5:22

$\DeclareMathOperator{\Int}{Int}\DeclareMathOperator{\Bd}{Bd}$ The most interesting thing I came up with, which may be a special case of something more interesting, is that if $X$ is connected, $A\subset X$, and $\Int A$ and $\Bd A$ are connected, then $\bar{A}$ is connected:

Suppose that $B$ and $C$ form a separation of $\bar{A}$. Since $\bar{A}$ is closed, $B$ and $C$ are closed in $X$. Since $\bar{A}=\Int A\cup\Bd A$ and the interior and boundary are connected, we can assume without loss of generality that $\Int A\subset B$ and $\Bd A\subset C$, so in fact $\Int A=B$. Furthermore, since $\Int A$ is connected, so is its closure, and $\overline{\Int A}\subset\bar{A}$, so $\overline{\Int A}=\Int A$. But this contradicts the fact that $X$ is connected.

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