# A connected set that must intersect the boundary of an open set given certain conditions

Suppose that $C$ is a connected subset of $X$, and $U$ an open set in $X$. Also, we have that $C$ contains a point in $X$ and a point not in $X$. Then how do we show that the following is true: $$C \cap \partial U \neq \varnothing?$$

My second question is that what is a good example in $\mathbb{R}^2$ of a connected set $C$ with an open set (disk) $U$ such that $C$ contains a point inside $U$ and a point in the complement of cl$(U)$, i.e. the closure of $U$, and some component of $U \cap C$ misses $\partial U$?

For the first question, right now the only thing that comes to mind is a proposition that states for $A \subset X$, if $C$ is a connected subspace of $X$ such that $$C \cap A \neq \varnothing, C \cap (X-A) \neq \varnothing,$$ then $C \cap \partial A \neq \varnothing$. In this case, however, we don't know if $A$ is open, and nowhere is it mentioned that $C$ contains a point in $X$ and a point not in $X$. So how should we proceed? I would also really appreciate some guidance on the second question for a suitable example and why it works.

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C is a subset of X, how can it contain a point not in X? – Amr Nov 8 '12 at 0:48
I think you meant not in U – Amr Nov 8 '12 at 0:48

I assume that you meant to say that $C$ contains a point of $U$ and a point not in $U$.

HINT: Let $V=X\setminus\operatorname{cl}U$. Show that if $C\cap\operatorname{bdry}U=\varnothing$, then $U\cap C$ and $V\cap C$ form a separation of $C$, contradicting the assumption that $C$ is connected.

For the second question practically any example will work. For instance, let

$$U=\big\{\langle x,y\rangle\in\Bbb R^2:x^2+y^2<1\big\}\;,$$

the open unit disk centred at the origin, and let $C$ be the $x$-axis. The boundary of $U$ is the unit circle, and the only component of $U\cap C$ is $\big\{\langle x,0\rangle:-1<x<1\big\}$.

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What about the second question? I think you might be right in regard to the first question with the confusion between $X$ and $U$. Was it even necessary for $U$ to be open? – Libertron Nov 8 '12 at 0:59
@Sachin: Yes, you need $U$ to be open. Related results can be stated for non-open $U$, but they’re more complicated. I’ve already added an example for the second question, but I’m not sure why it’s a problem: if you’ve stated it correctly, it’s hard not to find an example! – Brian M. Scott Nov 8 '12 at 1:03
haha, actually the example you gave was one I was playing around with to begin with! – Libertron Nov 8 '12 at 1:16

I will assume that you meant "another point not in $U$", I will call this point $y$. If $y \in \partial U$ then we are done. Therefore, assume $y \notin ∂U$ and assume $C∩∂U≠ \varnothing$ . Now Let $V$ be the interior of the complement of $U$. Now we have $x \in C ∩ U$ and $y \in C ∩ V$ since $C$ is a subset of the union of the open disjoint sets $U$ and $V$ (because $C∩∂U≠ \varnothing$), thus we get a contradiction because $C$ is connected.

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Umm… Aren't we supposed to assume that $C \cap \partial U = \varnothing$, if we are to do a proof by contradiction? – Libertron Nov 8 '12 at 1:14

I would suggest to do as follows: Take A=int(U)∩C and B=int(complement of U)∩C. We know that A∩B = ∅. Since C is not a subset of A, c is not a subset of A∪B. Thus C≠A∪B. By taking the intersection of both sides with ∂U we can show the latter.

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