I'm currently working through "Tensor Analysis on Manifolds" by Richard L. Bishop and Samuel I. Goldberg and came across proposition (a) which I uploaded here. It's the proof's last sentence that caught my eye. On the one hand it appeared intuitive that every neighborhood of a point x of a subset's closure intersects the subset. On the other hand I wanted proof.

So I went back a few pages to look whether I had missed something and, indeed, I had: On this page, the authors give an alternative definition of closed sets as well as the closure and the boundary of a set in case a basis of neighborhoods of the topological space X is given. Still, as all of this belongs to a short recap of topology, they gave no proof.

…which is why I tried to proof it myself. First of all, Bishop and Goldberg defined open and closed sets as follows: Open sets are given by the topology T, closed ones are the respective complements of the first. The interior and the closure of a subset A are:

$$\begin{align} A^0 &= \bigcup \text{All open sets contained in}~ A \\ \overline{A} &= \bigcap \text{All closed sets containing}~ A \end{align}$$

While I had no trouble showing the statement about closed sets, my proof of the equivalence of above's definition for $\bar{A}$ and the alternative one ("The closure of A consists of those x such that every basis neighborhood of x intersects A") went like this:

"$\subset$": Let $x \in A^0$ (as defined above) and let U be a basis neighborhood. Show that $U \cap A \neq \emptyset$.

Assume $U \cap A = \emptyset \Rightarrow U \subset X-A$. W.l.o.g let U be open (otherwise take $U^0$ instead) $\Rightarrow X-U$ is closed. Also, since $U \cap A = \emptyset$, it follows that $A \subset X-U$, which contradicts the fact that $x$ is in every closed superset of A.

The part I'm anything but sure about is the "w.l.o.g let $U$ be open". Again, it appeared intuitive in the first moment but, thinking about it again and considering that the topology is induced by the basis of neighborhoods (which, as I understand, doesn't necessarily consist of open neighborhoods), I couldn't come up with an explanation that $U^0 \neq \emptyset$ (let alone $x \in U^0$) if I don't assume that $X$ is Hausdorff (like in proposition (a)).

But maybe I'm on the wrong path? Anyway, I'd be glad if someone of you could shed some light on that matter.

  • 1
    $\begingroup$ In your definition of closure, shouldn't it be $A\subset B$ instead of $B\subset A$? $\endgroup$ – Jason DeVito Mar 29 '12 at 14:00
  • $\begingroup$ Yes, you're right. Damn copy&paste. $\endgroup$ – balu Mar 29 '12 at 20:19

In my (Dover) edition of Bishop and Goldberg's text, they define a neighborhood as follows (page 10, second paragraph)

A neighborhood of $x\in X$ is any $A\subset X$ such that $x\in A^0$.

So, your "wlog" is completely fine. If $U\cap A = \emptyset$, then $U^0\cap A = \emptyset$ as well since you've shrunk $U$ and $x\in U^0$ by definition of neighborhood.

  • $\begingroup$ Yes, in my (Dover) edition they define it that way, too. ;) Thanks for your help! That requirement that a basis neighborhood of some point x must always contain x as interior point must have slipped past me somehow. $\endgroup$ – balu Mar 29 '12 at 20:33

I am very confused by your question. Are you asking to prove that the following are equivalent?

$\textbf{P}$: $x \in X$ is in the closure $\bar{A}$ of a set

$\textbf{Q}$: Every neighbourhood $U$ about $x$ intersects $A$?

$\underline{\neg Q \implies \neg P}$

Let $V$ be an open set about $x$ that does not intersect $A$. Then $X - V$ is a closed set that contains $A$ with $x \notin X - V$. But then $\bar{A} \subset X - V$ by definition of the closure. Therefore by the contrapositive again, $(X-V)^{c} \subset (\overline{A})^{c}$. However by construction of $X-V$, $x \notin (X - V)$ so that $x \in (\overline{A})^{c}$.

$\underline{\neg P \Longleftarrow \neg Q}$

Suppose $x$ is not in the closure of $\overline{A}$. Then there is a closed set $V$ that contains $A$ such that $x \notin V$. But then this means that $X - V$ is an open set that contains $x$ that is disjoint from $A$.

Consequently we have shown that there is a neighbourhood about $x$ that does not intersect $A$.

$\hspace{6in} \square$

  • $\begingroup$ Yes, I know about the contrapositive. But as far as I can see you're not proving "that a point x is in the closure $\bar{A}$ of a set if every neighbourhood U about x intersects A" – which would be $A \Rightarrow B$ or, as contrapositive, $\neg B \Rightarrow \neg A$ (A being the statement that all neighborhoods intersect the subset) –, but the opposite ($B \Rightarrow A$ or $\neg A \Rightarrow \neg B$). However, my question's intent was slightly different, anyway. You implicitly assumed this neighborhood V not intersecting with the subset to be an open set. $\endgroup$ – balu Mar 29 '12 at 20:52
  • $\begingroup$ Anyhow, thanks for your proof. It's nice to see all this working without needing to introduce basis neighborhoods. $\endgroup$ – balu Mar 29 '12 at 20:53
  • $\begingroup$ @codethief Sorry I actually proved the converse of what you asked. I will edit my answer to put in a proof of the direction that you actually asked for. $\endgroup$ – user38268 Mar 29 '12 at 21:10
  • $\begingroup$ @codethief In future perhaps you should make your question a little clearer, I had difficulty trying to understand what it is you did not understand. $\endgroup$ – user38268 Mar 29 '12 at 21:20

I believe it comes down to noticing the following:

  • If $\mathcal{B}$ is a basis of neighbourhoods of $x$, then by definition $x \in U^\circ$ for all $U \in \mathcal{B}$.

  • If $\mathcal{B}$ is a basis of neighbourhoods of $x$, then so is $\mathcal{B}_0 = \{ U^\circ : U \in \mathcal{B} \}$.

  • If $\mathcal{B}, \mathcal{D}$ are two bases of neighbourhoods of $x$, then $U \cap A \neq \emptyset$ for all $U \in \mathcal{B}$ iff $V \cap A \neq \emptyset$ for all $V \in \mathcal{D}$.


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