Let $B \subseteq A \subseteq \mathbb{R}^n$. Show that $B$ is closed relative to $A$ iff $B = A \cap C$ for some set $C$ closed in $\mathbb{R}^n$.

My solution:

We know $B$ is closed relative to $A$ if $\forall x \in A \setminus B$, $\exists$ a neighborhood $N$ such that $N \cap B = \varnothing$. So, for each $x \in B \setminus A$, we have a nghbd $N_x = D^n(x,r_x) \cap A$ such that $N_x \cap B = \varnothing$. Put $O = \bigcup_{x \in A \setminus B} D^n(x,r_x)$. Obviously $O$ is open since it is a union of open discs. So $C = \mathbb{R}^n \setminus O$ must be closed. Now, since $B$ is closed in $A$, then $A \setminus B$ must be open in $A$, therefore $A \setminus B = O \cap A$. I have here a question since I cannot see how to conclude from here that $B = C \cap A$. Can someone help see it?

the other direction: Say $B = C \cap A$ for some $C$ closed in $\mathbb{R}^n$. Therefore, every $x \in A \setminus B$ must satisfy $x \notin C$. Since $C$ is closed, $\exists$ a disc $D^n(x,r)$ such that $x \in D^n(x,r)$ and $D^n(x,r) \cap C = \varnothing$. Put $N = D^n(x,r) \cap A$. Then, $ x \in N$ and $N \cap B \subseteq N \cap C$. Therefore, $B$ is closed in $A$.

Is this proof correct? any feedback? thanks

  • 1
    $\begingroup$ Please, make titles more informative! $\endgroup$ – Pedro Tamaroff Sep 2 '13 at 2:58

You’re doing fine up through the point at which you define $C$. At that point, however, you’re not entitled to conclude that $A\setminus B=O\cap A$; the fact that $A\setminus B$ is open in $A$ tells you that it’s equal to $U\cap A$ for some open $U$ in $\Bbb R^n$, but you’ve not actually shown that we can take this $U$ to be $O$.

It’s not hard to do so, however. Recall that $O=\bigcup_{x\in A\setminus B}N_x$, where $x\in N_x$ and $N_x\cap B=\varnothing$. On the one hand for each $x\in A\setminus B$ we have $x\in N_x\subseteq O$, so $A\setminus B\subseteq O$; and since $A\setminus B\subseteq A$, we actually have $A\setminus B\subseteq O\cap A$. On the other hand, for each $x\in A\setminus B$ we have $N_x\cap A\subseteq A\setminus B$, so that $$O\cap A=\left(\bigcup_{x\in A\setminus B}N_x\right)\cap A=\bigcup_{x\in A\setminus B}(N_x\cap A)\subseteq A\setminus B\;.$$ Thus, $O\cap A$ is indeed $A\setminus B$: your conclusion was premature, but none the less correct.

From here it’s just a bit of set algebra to see that $B=C\cap A$:

$$C\cap A=(\Bbb R^n\setminus O)\cap A=A\setminus O=A\setminus(O\cap A)=A\setminus(A\setminus B)=B\;.$$

You should verify each of those steps, since that kind of manipulation can be quite useful. However, the actual idea involved here is much simpler: $O$ and $C$ are complementary subsets of $\Bbb R^n$, so for any $X\subseteq\Bbb R^n$, $O\cap X$ and $C\cap X$ must be complementary subsets of $X$. Every point of $\Bbb R^n$ is in exactly one of $O$ and $C$, so clearly every point of $X$ is in exactly one of $O\cap X$ and $C\cap X$. Since $O\cap A$ is $A\setminus B$, $C\cap A$ must be all the rest of $A$, which is simply $B$.

  • $\begingroup$ @Citizen: You’re very welcome! $\endgroup$ – Brian M. Scott Sep 2 '13 at 3:45

I think your second proof is correct.
As to the first, you only need to notice that, if $x\in(C\cap A),$ then $x$ cannot belong to $O$. That is, $x\in A\backslash A\cap O=A\backslash(A\backslash B)=B,$ hence $C\cap A\subset B.$ The other direction is similar: $B\subset A\backslash(A\backslash B)=A\backslash(A\cap O)=A\backslash O=A\cap C.$
Hope this helps.


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.