topology problem and proof

Question

Let $X \not= \emptyset$ and $(X,\tau)$ be a topological space,$H \subseteq X$ and $G \subseteq X$. Show $\overline{H} \setminus \overline{G}\subseteq \overline{H \setminus G } $

This looks to me like an exercise in juggling definitions; you have closures, set differences and subsets. Have you tried applying the definitions directly? — Arthur, May 27 '20 at 12:35
Does this answer your question? Difference of closures and closure of difference — Rick, May 27 '20 at 12:56

score 1 · Answer 1 · answered May 27 '20 at 12:57

You may write $H$ as $$H = (H\setminus G)\cup (H\cap G)$$ so that $$\overline{H} = (\overline{H\setminus G})\cup (\overline{H\cap G})$$ Then \begin{align*} \overline{H}\setminus\overline{G}&=((\overline{H\setminus G})\cup (\overline{H\cap G}))\setminus ((\overline{G\setminus H})\cup (\overline{H\cap G}))\nonumber\\ &=(((\overline{H\setminus G})\cup (\overline{H\cap G}))\setminus (\overline{G\setminus H}))\cap (((\overline{H\setminus G})\cup (\overline{H\cap G})) \setminus(\overline{H\cap G}))\nonumber\\ &=(\overline{H}\setminus (\overline{G\setminus H}))\cap(\overline{H\setminus G})\nonumber\\ &\subseteq \overline{H\setminus G} \nonumber \end{align*}

score 0 · Answer 2 · answered May 27 '20 at 12:55

Let $x \in \overline{H}\setminus\overline{G}$. This means that every open set $U$ containing $x$ intersects $H$ but there exists an open set $V$ containing $x$ that does not intersect $G$. (By "$A$ intersects $B$", I mean that $A\cap B \neq \emptyset$.)
Fix $V$ as above.

We wish to show that $x \in \overline{H\setminus G}$. Let $U$ be an open set containing $x$.
Then, we have that $U\cap H \neq \emptyset$. (Why?)

Claim. $U \cap H \not\subset G$.
Proof. Suppose not. Then, we would have $U \cap H \subset G$.
We would also have $V\cap(U\cap H) \subset V\cap G$. However, note that $V \cap G = \emptyset$.
Thus, we get that $V\cap(U\cap H) = \emptyset$ or $(V \cap U) \cap H = \emptyset$.
However, note that $V \cap U$ is an open set containing $x$ (why?) and thus, the above contradicts that $x \in \overline H$. (Why?)
This proves our claim.

Thus, we have shown that $U\cap H\neq\emptyset$ and $U \cap H \not\subset G$. This gives us that $$U \cap (H \setminus G) \neq \emptyset.$$ (Why?)

As $U$ was an arbitrary open set containing $x$, we are done.

Note that I haven't assumed $G \subset H$. Also, make sure that you understand the definitions of closure and complements well. Else, you would have trouble understanding the above.

topology problem and proof

2 Answers2