Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
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. –  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 $A$.

$\hspace{6in} \square$

share|improve this answer
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. –  balu Mar 29 '12 at 20:52
Anyhow, thanks for your proof. It's nice to see all this working without needing to introduce basis neighborhoods. –  balu Mar 29 '12 at 20:53
@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. –  user38268 Mar 29 '12 at 21:10
@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. –  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 $V \in \mathcal{D}$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.