Question about proof from Bishop & Goldberg

[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.

 

[fresh_42]

"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."

\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.

 

[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.

 

[fresh_42]

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.

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.

 

[TeethWhitener]

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 \subseteq X_0. So we usually may substitute X_0 by U_0 and thus getting an open neighborhood.

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.

 

[fresh_42]

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.

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).

 

[Cruz Martinez]

Thanks for the reply @fresh_42.
[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.

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.

 

[Cruz Martinez]

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).

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 $x \in B \subseteq U$.

 

[TeethWhitener]

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 $x \in B \subseteq U$.

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.

 

[zinq]

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

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

V = ∩_{a ∈ J} G_a^x​

will be an open set containing x that is disjoint from A. (Since each G_a^x is disjoint from the corresponding G_a.) 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.

 

[Cruz Martinez]

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.

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 doesn't 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.

 

[TeethWhitener]

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

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.