# >$V, W$ are open sets in $X$ with $V\subseteq W$ and $\partial V \cap W = \emptyset$. Then $V$ is a union of components of $W$.

This is statement 10.16 in Rudin's Functional Analysis. I just need help clarifying a statement.

$V, W$ are open sets in $X$ with $V\subseteq W$ and $\partial V \cap W = \emptyset$. Then $V$ is a union of components of $W$.

Proof: Let $\Omega$ be a component of $W$ that intersects $V$. Then as $V$open, $X = V \cup \partial V \cup \overline{V}^c$ and this is a disjoint union. As $\Omega \subseteq W$ and $W \cap \partial V = \emptyset$, then $\Omega = (\Omega \cap V) \cup (\Omega \cap \overline{V}^c)$, a disjoint union of open sets. As $\Omega$ is connected, $\Omega = (\Omega \cap V)$ that is $\Omega \subseteq V$ and the result follows.

So my problem is in seeing $\Omega \cap V$ and $\Omega \cap \overline{V}^c$ as open subsets of $X$ or $W$. If $X$ were locally connected, then there is no issue but that wasn't assumed in the text. I'm assuming the result would follow from openness of $W$ but I can't see how.

EDIT: So I think I've scrapped together a solution avoiding the openness problem. Could anyone verify if it is correct for me?

Viewing $V$ and $\Omega$ as sets in space $W$, we have $\partial_WV \subseteq \partial V \cap W$ so $\partial_W V = \emptyset$. $V$ is open in $W$ so $W = V \cup \partial _W V \cup W \setminus cl_W(V) = V \cup W \setminus cl_W(V)$, a disjoint union of open sets in $W$. $\Omega \subseteq W$ and $\Omega$ would be partitioned by $V$ and $W \setminus cl_W(V)$. As $\Omega$ connected and intersects $V$, then $\Omega \subseteq V$ as wanted.

They don't necessairily have to be open sets of $X$ nor $W$:

We say that the topological space $\Omega$ (viewed as a topological space itself, in this case, with the subspace topology of $X$) admits a disconnection if $\Omega=A \cup B$ with $A,B$ disjoint nonempty, open sets of $\Omega$. We say that a topological space is connected if it admits no disconnection.

In the present case, both of the sets $\Omega \cap V$ and $\Omega \cap \overline{V}^c$ are open in $\Omega$ since they are the intersection of an open set of $X$ with $\Omega$, which is precisely the condition we need, to use the hyphotesis of $\Omega$ being connected.

• Thanks, this was a really silly question in hindsight – E.Lim Oct 20 '17 at 6:43

Rudin means the disjiont open sets are open in $\Omega$, not necessarily in $W$ or $X$.

A subset $Y$ of a topological space $X$ is called disconnected if $Y$ is disconnected as a subspace of $X$ or there are two disjiont open subsets of $X$ with nonempty intersection with $Y$.

• I'm having trouble seeing how they satisfy the condition. Could you spell it out for me? Also is there a standard result showing how one boundary is a subset of another boundary? – E.Lim Oct 18 '17 at 14:30
• Sorry, the original answer is questionable. I have made an edit. – C.Ding Oct 19 '17 at 8:50
• Right, I was being silly, didn't think of $\Omega$ as a space itself. – E.Lim Oct 20 '17 at 6:41