# Question about proof from Bishop & Goldberg

1. Feb 20, 2016

### TeethWhitener

I'm going through Bishop and Goldberg's "Tensor Analysis on Manifolds" right now and I'm stuck in Chapter 0. They give a proof of the statement "A compact subset of a Hausdorff space is closed" that I can't seem to wrap my head around. I'm reprinting the proof here:

"Suppose that $A$ is a compact subset in a Hausdorff space $X$ and $A \neq A^-$ (where $A^-$ denotes the closure of $A$), so there is an $x \in A^- - A$. For every $a \in A$ there are open sets $G_a, G^x_a$ such that $G_a \cap G^x_a = \emptyset , a \in G_a$, and $x \in G^x_a$, because $X$ is Hausdorff. Then $\{G_a|a \in A\}$ is an open covering of $A$, so there is a finite subcovering $\{G_a|a \in J\}$, where $J$ is a finite subset of $A$."

I'm fine with everything up to this point, but the next sentence loses me:

"But then $\bigcap_{a \in J} G^x_a$ is a neighborhood of $x$ which does not meet $\bigcup_{a \in J} G_a \supset A$, so $x$ cannot be in $A^-$, a contradiction."

The authors have previously defined the closure of a set $A$ as the intersection of all closed sets containing $A$. I get that $\bigcap_{a \in J} G^x_a$ is not a subset of $A$, but I don't understand why that implies that $x \notin A^-$.

[EDIT]: I recalled another section earlier in the book saying that $x \in A^-$ iff every neighborhood of $x$ intersects $A$, which makes the last sentence in question trivial.

Last edited: Feb 20, 2016
2. Feb 20, 2016

### Staff: Mentor

$\bigcap_{a \in J} G^x_a$ is an open neighborhood of $x$ and $(\bigcap_{a' \in J} G^x_{a'}) \cap (\bigcup_{a \in J} G_a) = \bigcup_{a,a' \in J} (G^x_{a'} \cap G_a) = ∅$.
If $G^x_{a'} \cap G_a ≠ ∅$ for some pair $(a',a)$ we can replace $G^x_{a'}$ by $G^x_{a'} \cap G^x_{a}$. Remember $J$ being finite.
Thus we have for any $x∈X\setminus{A}$ an open neighborhood $\bigcap_{a \in J} G^x_{a}$ that does not intersect with $A$, i.e. any such $x$ is an inner point of $x∈X\setminus{A}$ which means $x∈X\setminus{A}$ is open and $A$ is closed.

3. Feb 21, 2016

### TeethWhitener

Thanks for the reply @fresh_42. Maybe you can help me out with one further question. The proof above relies on the fact that, according to the book, "$x \in A^-$ iff every neighborhood of $x$ intersects $A$." But I can only seem to show this for open neighborhoods.

For instance, let $x \in A^-$ and assume that there's a neighborhood $X_0$ of $x$ such that $X_0 \cap A = \emptyset$. If $X_0$ is open, then the complement $X^c_0$ is closed and $A \subset X^c_0$. The fact that $X^c_0$ is closed implies that $A^- \subset X^c_0$ as well. But this means that $X^c_0 \cap A^- = A^-, X_0 \cap A^- = \emptyset$ and therefore $x \notin A^-$.

This only works because we assumed that $X_0$ was open. Can we say something analogous about non-open neighborhoods? I get the impression from looking around that in general, "neighborhood" tends to refer to "open neighborhood," but in the book that I'm using, "neighborhood of $x$" is just defined as any set for which $x$ lies in the interior, without regard to whether that set is open/closed/both/neither.

But upon further reflection, I'm thinking that maybe this doesn't hold for non-open neighborhoods. If we consider $\mathbb{R}$ with the standard topology, then the singleton set $\{x\}$ is closed, as it's the complement of $(-\infty , x)\cup(x,\infty)$. If we take $\{x\}$ as a neighborhood of $x$, then it clearly can be in $A^-$ without intersecting $A$. But by the same token, $\{x\}$ doesn't seem to have an interior, so maybe it doesn't count as a neighborhood of $x$. Is this right?

[EDIT]: Just found that I misspoke above. The book actually says "$x \in A^-$ iff every basis neighborhood of $x$ intersects $A$." The paragraph before that mentions open sets as a basis of neighborhoods. I still can't really figure out the importance of a neighborhood versus a basis neighborhood, though.

4. Feb 21, 2016

### Staff: Mentor

What does this mean "in the book that I'm using, "neighborhood of $x$" is just defined as any set for which $x$ lies in the interior"? Let's say $x$ is in the interior of $X_0$ then this means there is a open neighbourhood $U_0$ of $x$ such that $U_0 ⊆ X_0$. So we usually may substitute $X_0$ by $U_0$ and thus getting an open neighborhood. If $x$ happens to be in $\overline{X_0} \setminus X_0$ then it will not work, e.g. $\{1\} ∈ [0,1] ∩ [1,2]$.
Your paragraph before the last is correct.

I don't know the definition of a basis neighbourhood, maybe you can help me with this.
Usually one examines (open) coveries containing a given set or the whole topological space.
In case of compactness this means there is a finite set of open coveries which makes it far easier to intersect or build unions of all closed or all open sets.

5. Feb 21, 2016

### TeethWhitener

Thank you, this finally cleared the logjam in my head. Just to be clear, it doesn't matter whether $X_0$ is open or not: As long as the interior of $X_0$ is nonempty, we can use it in the proof instead and it'll apply to the whole neighborhood as a superset of the interior. This is why singleton sets don't work: their interior is empty.

As for a basis of neighborhoods, the book I'm using doesn't give a great definition. Any open set in a topology can be expressed as a union of basis sets, so I assume a basis of neighborhoods is a collection of neighborhoods such that any other neighborhood can be expressed as a union of the basis neighborhoods. At any rate, the basis is defined in terms of open sets, and coupled with your insight above, I think everything is clearer for me now. Thanks again for your help.

6. Feb 21, 2016

### Staff: Mentor

You're welcome!
A topology is defined by its open sets. What you call basis I know as open cover (maybe coverage in English? but Wiki says cover).

7. Feb 23, 2016

### Cruz Martinez

Yes, $x \in \bar{A}$ iff every basis element containing $x$ intersects $A$. Note however that the topology itself is a basis (How can you write any open set as a union of elements in this basis?). So that automatically $x \in \bar{A}$ iff every neighborhood of $x$ intersects $A$.

Edit: Most authors use "neighborhood of $x$ to mean any open set $U$ which contains $x$, while some others define it as any set which contains an open set which contains $x$, so always check this in the book you're using. Usually very minor modifications in arguments/proofs are needed.

Last edited: Feb 23, 2016
8. Feb 23, 2016

### Cruz Martinez

No, both concepts are very different. What you describe can be defined as: A collection of open sets $U_{ \alpha}$ such that $A \subseteq \union_{ \alpha} U_{ \alpha}$ is an open covering of $A$. What the OP is describing is a basis, which is a collection of open subsets $\mathscr{B}$ such that for any open set $U$ and any $x \in U$, we can find $B \in \mathscr{B}$ such that $x \in B \subseteq U$.

9. Feb 23, 2016

### TeethWhitener

Thanks, I got the impression from my book that a basis and an open cover were quite different as well, and along the lines of what you said above.

10. Feb 23, 2016

### zinq

Referring to the original post: The final point is that you have obtained a finite covering {Ga, a in J} of the set A by open sets. And for each Ga of these finitely many open sets, there is another open set Gax containing x that is disjoint from Ga.

And since the finite intersection of open sets is open, and since the union of the finitely many sets {Ga, a in J} contains A, this guarantees that the intersection of the sets Gax:

V = a ∈ J Gax

will be an open set containing x that is disjoint from A. (Since each Gax is disjoint from the corresponding Ga.) And that's a contradiction, since by initially assuming that A was not closed we found the point x not in A but whose every open neighborhood (including V) had to intersect A.

11. Feb 23, 2016

### Cruz Martinez

Just as a note on this, the interior of the singleton $\{x\} \subseteq X$ is not always empty. What happens if you take the topological space $(X, \tau)$ where $\tau = \mathcal{P}(X)$?

And yes, both definitions of neightborhood give the same characterization of closure, namely: A point $x$ is in the closure of $A$ if and only if every neighborhood of $x$ has nonempty intersection with $A$. So it doesnt matter wether you think of a neighborhood as an open set containing $x$ or as a set containing an open set containing $x$, for *this* particular proof.

I will prove it here: Suppose every open neighborhood of $x$ intersects $A$. Let $V$ be a neighborhood of $A$, then $V$ contains an open set $U$ containing $x$, since $U \cap A \neq \emptyset$, then $V \cap A \neq \emptyset$ because $V$ contains $U$.
Conversely, suppose every neighborhood of $x$ (open or not) intersects $A$, then in particular all the open neighrborhoods intersect $A$.
Now you can easily prove that $x \in \bar{A}$ if and only if every open neighborhood of $x$ intersects $A$ and this will complete the argument that both definitions of neighborhood can be used interchangeably to characterize closure points.

12. Feb 23, 2016

### TeethWhitener

That's true. The post you quote was referring to an earlier post where I specifically mentioned singletons in $\mathbb{R}$ with the standard topology.

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