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This is my first post here as I come from a physics background and I finally decided that it is about time to delve into more formal mathematics.

Lately, I have been working through "Tensor Analysis on Manifolds" by Richard L. Bishop and Samuel I. Goldberg. In it, I was going through the Topology primer 0.2.1 section and noticed an interesting claim:

"A set $G$ is closed iff every basis neighbourhood of $x$ intersects $G$, then $x \in G$".

How do I formally prove it? The statement does not even sound intuitive for me. I'd be very glad is somebody shed some light on this :)

It is also worth to mention that my question is related to another question. Sadly, it only found the later claims question worthy and not the above one.

Kosutusas
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  • Hi! @Kosutusas. Welcome Math. S.E, maybe can you show your attempts? –  Aug 20 '20 at 15:46
  • What definition of closed are you working from? In my mind, the claim you give is actually the definition of closed sets. – jben2021 Aug 20 '20 at 16:11
  • hey @Ramanujan ! I'd love to, but I am a bit confused as to how to even start the proof as I don't believe I understand the claim. Best I can do is to attempt to write the claim formally.

    We take $\mathcal{B}(x)$ as the basis of neighbourhoods around x, then: $U \cap G \neq \emptyset, \forall U \in \mathcal{B}(x) \Leftrightarrow G$ is a closed set with $x \in G$.

     – Kosutusas Aug 20 '20 at 16:12
  • hey @jben2021! My definition of closed is the complement of open sets. While open sets are elements of the topology. – Kosutusas Aug 20 '20 at 16:14
  • Basically it's saying that a set $A$ is closed if and only if it satisfies the following separation property: if there's no neighborhood of $x$ not intersecting $A$, then $x\in A$. When you prove these "if and only if" claims, you take one direction at a time. Let's assume first that your set is closed so that its complement is open. Can you see how to show that this implies that separation property? – jben2021 Aug 20 '20 at 16:16
  • It may also help to think about open sets as "measuring distance" in some sense. The intuition is that if there is no neighborhood of $x$ not intersecting $A$ then you could say that $x$ is arbitrarily close to $A$. With this in mind, $A$ is closed if and only if "arbitrarily close to $A$" is equivalent to "in A." Hope that helps – jben2021 Aug 20 '20 at 16:21
  • Thanks, @jben2021, for the help, I really appreciate it :)

    Regarding the one direction of the proof, are you referring to the fact that if $G$ is closed and if $x \not\in G$, then $X-G$ is a neighbourhood of $x$ which (by definition of complement) does not intersect $G$. However, I don't see how for the case of $x \in G$, $X-G$ can imply anything about the neighbourhoods of x since it does not contain x.

     – Kosutusas Aug 20 '20 at 17:09
  • Yes, exactly! What you've shown is that $x\notin G$ implies that there is a neighborhood of $x$ not intersecting $G$. Then the contrapositive of this statement (contrapositive of A implies B is not B implies not A) is that if every neighborhood of $x$ intersects $G$, then $x\in G$, which gives one direction. Now prove the converse. – jben2021 Aug 20 '20 at 17:19
  • I gave the other direction as an answer in case you need it. But I encourage you to prove it yourself first. It looks like you're on the right track! – jben2021 Aug 20 '20 at 17:24

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I'm going to give the other direction to the proof in case you need it, @Kosutusas. Suppose that if $x$ admits no neighborhood not intersecting $G$ then $x\in G$. Then the complement of $G$ is such that every $x\in X-G$ has an open neighborhood also lying in $X-G$ (since we can find a neighborhood not intersecting $G$, hence lying in the complement). Therefore, $X-G$ is open since every point in $X-G$ admits a neighborhood lying entirely in $X-G$.

jben2021
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  • Thanks @jben2021. I do have a questions tho. You stated that:

    "Suppose that if x admits no neighborhood not intersecting $G$ then $x \in G$".

    Why does $\forall U \cap G \neq \emptyset$ imply $x \in G$ ($U$ is a neighborhood of x)? I would understand if it was the other way around via contrapositive of if $\exists U \cap G = \emptyset $ then $x \not\in G$.

     – Kosutusas Aug 21 '20 at 13:38
  • I was giving the other direction of the proof. You showed that A implies B in your comments above, and I was giving B implies A. You're totally right that this condition isn't always true, but we're trying to show that when we have this condition, we can conclude that our set is closed. Hope that clears it up – jben2021 Aug 21 '20 at 15:16