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My first exposure to any sort of topology is from Spivak's Calculus on Manifolds. I think I understand compactness conceptually, I'm just finding the rigor a little bit elusive.

My first question is regarding a bit of notation. I'm really not sure what $\{ x\} \times B $ is saying

The theorem in question is

Theorem: If $B$ is compact and $\mathcal O$ is an open cover of $\{x \} \times B$, then there is an open set $U \subset \mathbb{R}^n$ containing $x$ such that $U \times B$ is covered by finite number of sets in $\mathcal O.$

My second question is about the following proof:

(Original screenshot)

Proof. Since $\{x\}\times B$ is compact, we can assume at the outset that $\def\O{\mathcal O}\O$ is finite, and we need only find the open set $U$ such that $U\times B$ is covered by $\O$.

For each $y\in B$ the point $(x,y)$ is in some open set $W$ in $\O$. Since $W$ is open, we have $(x,y)\in U_y\times V_y\subset W$ for some open rectangle $U_y\times V_y$. The set $V_y$ cover the compact set $B$, so a finite number $V_{y_1}, \ldots V_{y_k}$ also cover $B$. let $U=U_{y_1}\cap\cdots\cap U_{y_k}$. Then if ($x',y')\in U\times B$, we have $y'\in V_{y_i}$ for some $i$ (Figure 1-4), and certainly $x'\in U_{y_i}$. Hence $(x',y')\in U_{y_i}\times V_{y_i}$, which is contained in some $W$ in $\O$.

It's a bit difficult to understand the proof without knowing what $\{ x\} $ is, but it would appear to me that it is a set containing a point $x$.

The dimensionality of the subject has me confused. We're saying that "for each $y \in B$, there is a point $(x,y)$ in some open set $W$. Why have we picked a two dimensional point instead of something more general? The figure used to illustrate the problem depicts a set in $\mathbb R ^2$ as well. The very name used "open rectangle" confuses me, it feels as though we're restricted to only describing things in two dimensional "slices".

We next say that $(x,y)\in U_y \times V_y \subset W$ for some open rectangle. How do we know that the sets $V_y$ cover $B$ (and also why are we using a point in $\mathbb R ^2$ )? I think I understand the rest of the proof, but this bit (which is the most important part) isn't really making sense to me.

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The Cartesian product of sets $A$ and $B$ is $A \times B = \left\{ (a, b) \; \mid \; a \in A, \; b \in B \right\}$, so a product with a singleton set is just a copy of the other set. – Sammy Black Feb 24 '14 at 21:57
Is that the proof for $A \times B$ is compact if both $A$ and $B$ are compact? – dani_s Feb 24 '14 at 21:59
I looked up in my book and found the theorem. Theorem : If $B$ is compact and $\mathcal O$ is an open cover of $\{x \} \times B$, then there is an open set $U \subset \mathbb{R}^n$ containing $x$ such that $U \times B$ is covered by finite number of sets in $\mathcal O$ – Nameless Feb 24 '14 at 23:08
Oh wow, sorry. I thought that I included that in the screenshot. I'm sorry about that. – Astrum Feb 24 '14 at 23:11
up vote 4 down vote accepted
  1. Yes, $\{x\}$ is a set which has a single element (and thus we call it a singleton), and $\{x\}\times B=\{(x,b)\mid b\in B\}$. Clearly it is homeomorphic to $B$.

  2. Note that ordered pairs are not necessarily ordered pairs of real numbers. You can talk about product of two sets, or two spaces. And this is somewhat similar to the case of $\Bbb R^2$, or the real plane, being the product of two copies of $\Bbb R$.

    Open rectangles, if so, refers to the product of two open sets. So $U\times V$ is an open rectangle in the product $X\times Y$ if $U\subseteq X$ and $V\subseteq Y$ are both open in the respective spaces.

  3. The sets $V_y$ cover $B$ because for every $y\in B$ we chose some $V_y$ such that $y\in V_y$. Therefore the union of the $V_y$'s includes all the points from $B$, and therefore it covers $B$.

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It's a little difficult to give you any really useful help because you showed the proof of the theorem without saying what the theorem was.

$\{x\}$ is the set that contains the point $x$ and nothing else. (Similarly, $\{x, y\}$ is the set that contains $x, y, $ and nothing else.)

$A\times B$ generally is the set of all ordered pairs $(a,b)$ where $a$ is in $A$ and $b$ is in $B$. (So for example $R\times R$ is the set of all ordered pairs of real numbers, which we usually identify with the plane.) In this case, $\{x\}\times B$ is the set of all ordered pairs $(x, b)$ where $b$ is some element of $B$. This is the special case of the foregoing where $A$ has only one point.

A point of $A\times B$ always has exactly two coordinates, one from the $A$ space and one from the $B$ space, and it has a unique representation in the form $(a,b)$ where $a$ is in $A$ and $b$ is in $B$.

In topology we usually (not always) give $A\times B$ a special topology called the product topology; the statement of the theorem will probably mention whether the product topology is the one it wants, and will certainly mention it if it isn't, so if it doesn't say, you use the product topology. For infinite products the product topology is a little tricky, but here we have a finite product (of 2 sets) which is not tricky at all. if $G_A$ is open in $A$ and $G_B$ is open in $B$, then $G_A\times G_B$ is an open set of $A\times B$ in the product topology, and open sets of $A\times B$ are arbitrary unions of these. (That is, these sets for a basis for the topology.) For $R^2$ considered as $R\times R$, this construction gives the usual topology on $R^2$: every open set in $R^2$ is a union of open rectangles. By analogy, such products are called “open rectangles” even in product spaces that are completely unlike $R\times R$.

The most important thing about a product is that it has two projection functions back into the original sets. There is a function $\pi_A : (A\times B) \to A$ which takes $(a,b) \mapsto a$, and similarly $\pi_B$ takes $(a,b)\to b$. The product topology has the important property that it is the coarsest topology for which these projection maps are continuous. So in particular, projection maps are continuous!

The diagram is probably not trying to depict $R^2$. It is trying to depict $A\times B$ for arbitrary $A$ and $B$, but it looks like $R^2$ (which is an important special case) because that's how paper works.

I think the proof is trying to show that a product of two compact sets is compact. The idea is that if we have an open cover of $A\times B$ then we can take each of the open sets in the cover and project them, via $\pi_A$, back into $A$, and they must cover $A$. Then because $A$ is compact, we can chose a finite subcover of these image sets in $A$. Then we can lift this finite family of open sets of $A$ back to a finite subcovering of the original space $A\times B$.

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This answer and the one above were exactly what I was looking for. The proof was showing that the product of a singleton and a compact set is compact. The very next theorem in the book is showing that the product of any two compact sets is compact. – Astrum Feb 24 '14 at 22:35

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