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I am having a real struggle proving a property to do with the closure of a subset M $\subset$ X.

I understand the closure of a set M [; \subset ;] X to be: the smallest closed subset that contains M. (also, the intersection of all closed subsets containing M)

The property I am having trouble proving is:

$$ x \in \overline{M} \iff \forall \epsilon > 0, B(x,\epsilon) \cap \text{M} \neq \emptyset $$

I can prove this in one direction by proving the contrapositive, showing if $ x \notin \overline{\text{M}} $, then there exists a ball $ B(x,\epsilon)$ such that $ B(x,\epsilon ) \cap \overline{\text{M}} = \emptyset $ (since $ \overline{\text{M}} $ is closed). And since $ M \subset \overline{M}, B(x,\epsilon ) \cap M = \emptyset $

However, I am completely stumped how to prove the other direction. Here is a link to my class notes I describing the proof.

I would greatly appreciate any help you could offer me.

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You can show again the contrapositive: assume $\exists \epsilon_0 > 0: B(x, \epsilon_0) \cap M = \emptyset$. can you proceed showing that $x \notin \overline{M}$? (Hint: you need to show that $\exists F \subset X$ which is closed and $M \subset F$ but $x \notin F$)

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  • $\begingroup$ Thank you! Could you take F to be the complement of B(x,e). This contains M (because the intersection between the B(x,e) and M is the empty set) and is closed because B(x,e) is open. So then F is in the closure of M, and F does not contain x! $\endgroup$ – emidude Apr 21 '17 at 15:52
  • $\begingroup$ Oh wait. F is not in the closure of M. The closure of M is a subset of F. But if x is not in F, then x cannot be in the intersection of all the closed subsets containing M. So x is not in the closure of M. $\endgroup$ – emidude Apr 21 '17 at 16:14
  • $\begingroup$ Yes. That's what I thought of :) $\endgroup$ – AsafHaas Apr 22 '17 at 18:00
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Suppose $x \in \overline{M}$ (as defined by $$\overline{M} = \bigcap \{F: F \text{ closed } M \subseteq F\}$$

as your notes describe. Suppose that $x \in \overline{M}$ and let $r>0$. We want to show that $B(x,r) \cap M \neq \emptyset$. Suppose it does not. Then $B(x,r) \cap M = \emptyset$ so $M \subseteq X\setminus B(x,r)$. The latter is a closed set (I assume you know that "open balls" like $B(x,r)$ are indeed open sets, as their name suggests, and so their complement is open). As $\overline{M}$ is the intersection of all closed sets that contain $M$, it follows that $\overline{M} \subseteq X \setminus B(x,r)$ as well. But $x \in \overline{M}$ and $x \notin X \setminus B(x,r)$. This contradiction shows that in fact $$\text{(1) } \forall r>0: B(x,r) \cap M \neq \emptyset$$

Now suppose that $x$ satisfies the above condition. We want to see that $x \in \overline{M}$. So suppose again that is is not in the closure. This means that there exists some closed set $F$ such that

$$M \subseteq F \text{ and } x \notin F$$

This means that $x \in X \setminus F$ and the latter set is open, so by the definition of the metric topology

$$\exists r>0 : B(x,r) \subseteq X \setminus F$$

But then $$B(x,r) \cap M \subseteq B(x,r) \cap F =\emptyset$$

which contradicts the condition (1) on $x$.

So your condition that $M$ contains all its adherence points (which is what points satisfying (1) are called) is equivalent that $M$ is closed.

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