4
$\begingroup$

Old qual question here.

Let $X$ be connected and $C\subset X$ be closed such that the boundary of $C$ is a point. Show that $C$ is connected.

Attempt:

For contradiction, suppose $U,V$ are nonempty disjoint sets open in $C$ such that $C=U\cup V$. Call $y=\partial C$ and say $y\in U$. Then $V\subset\operatorname{int}C$, which is open, so $V$ is open in $X$. Since $C$ is closed in $X$, $C^C$ is open in $X$.

This is where I'm stuck. I think that $U\cup C^C$ has to be open in $X$, then $X=(U\cup C^C)\cup V$ would not be connected, a contradiction. But I'm stuck here.

$\endgroup$
2
  • $\begingroup$ A set with empty boundary is closed and open. $\endgroup$ – Rene Schipperus Jan 5 '16 at 2:01
  • $\begingroup$ @ReneSchipperus I see why that is true, but I am having trouble seeing why that solves the problem. $\endgroup$ – Logan Tatham Jan 5 '16 at 5:05
2
$\begingroup$

Note that since $C$ is closed, then $\partial C = C \setminus \operatorname{Int} (C)$, so $C = \{ x \} \cup \operatorname{Int}(C)$ where $\partial C = \{ x \}$.

Suppose that $C$ is not connected, so that there are $U,V$ open subsets of $X$ such that

  • $U \cap C \neq \emptyset \neq V \cap C$,
  • $C \subseteq U \cup V$,
  • $(U \cap V ) \cap C = \emptyset$.

Without loss of generality, $x \in U$. Then as $x \notin V$ by replacing $V$ with $V \cap C = V \cap \operatorname{Int} ( C )$ we may assume that $V \subseteq C$, and so in particular $V \cap U = \emptyset$.

Consider $U \cup ( X \setminus C )$, and $V$.

  • Clearly $U \cup ( X \setminus C)$ and $V$ are open subsets of $X$.
  • Note that $( U \cup ( X \setminus C ) ) \cup V = ( U \cup V ) \cup ( X \setminus C ) \supseteq C \cup ( X \setminus C ) = X$.
  • Note that $( U \cap ( X \setminus C ) ) \cap V = ( U \cap V ) \cup ( ( X \setminus C ) \cap V ) = \emptyset$.

Therefore $U \cup ( X \setminus C )$ and $V$ form a separation of $X$, contradicting that $X$ is connected! Thus it must be that $C$ is connected.

$\endgroup$
1
$\begingroup$

Here’s a slightly simpler argument (along the same general lines).

Suppose that $p$ is the unique boundary point of $C$, and note that $\operatorname{int}C=C\setminus\{p\}$. If $C$ is not connected, there are non-empty sets $H$ and $K$, both relatively closed in $C$, such that $H\cap K=\varnothing$ and $H\cup K=C$. $C$ is closed, so in fact $H$ and $K$ are closed in $X$. Without loss of generality assume that $p\in K$. Then $H=(\operatorname{int}C)\setminus K$, which is open, so $H$ is clopen (i.e., both closed and open). Since $\varnothing\ne H\ne X$, it follows immediately that $X$ is not connected: $H$ and $X\setminus H$ form a separation of $X$.

(If you don’t immediately see why $(\operatorname{int}C)\setminus K$ is open, note that it equals $(\operatorname{int}C)\cap(X\setminus K)$, the intersection of two open sets in $X$.)

$\endgroup$
1
$\begingroup$

Let $\partial C=\{p\}$, and let $V,W$ be open subsets of $X$ such that $C\subseteq V\cup W$ and $C\cap V\cap W=\emptyset$. We assume WLOG that $p\in V$, and our objective is to show that $B=\emptyset$, where $B:=C\cap W$.

If $A:=C^c\cup V$ then $A$ is open and nonempty ($p\in A$), $X=A\cup B$ and $A\cap B=\emptyset$, so by connectedness of $X$ it is suffices to show that $B$ is open. Now from $C=\operatorname{int}(C)\cup\{p\}, p\in V$ and $C\cap V\cap W=\emptyset\,$ we obtain $B\subseteq\operatorname{int}(C)$, so $B=\operatorname{int}(C)\cap W$, as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.