If $B$ is compact, then $\{x\}\times B$ is compact

Question

In Spivak's Calculus on Manifolds, he writes on page 8:

If $B\subset\mathbb R^m$ is compact, and $x\in\mathbb R^n$, then it is easy to see that $\{x\}\times B\subset\mathbb R^{n+m}$ is compact.

Here, $\{x\}\times B$ is defined as $\{(x^1,\dots,x^n,b^1,\dots,b^m):(b^1,\dots,b^m)\in B\}$. Spivak uses the above fact to prove that the product of two compact sets is compact, and so we are not allowed use this more general statement to conclude the above. Despite Spivak's assertion that this is "easy to see", I don't think the proof is completely trivial (or at least it is not short). Is my attempt correct?

The basic idea is that if $\{A_i\}_{i\in I}$ is an open cover of $\{x\}\times B$, then for each collection $A_i$ of open sets of $\mathbb R^{n+m}$, we can find a corresponding collection $\alpha_i$ of open sets of $\mathbb R^m$; from there, we can use the compactness of $B$ to find a finite subcover of $\{A_i\}_{i\in I}$.

In more detail, suppose $\{A_i\}_{i\in I}$ is an open cover of $\{x\}\times B$. For each $A_i\subset \mathbb R^{n+m}$, define $\alpha_i\subset\mathbb R^m$ so that $(y^1,\dots,y^m)\in\alpha_i$ if and only if $(x^1,\dots,x^n,y^1\dots,y^m)\in A_i$.

We claim that $\{\alpha_i\}_{i\in I}$ is an open cover of $B$. Indeed, if $(b^1,\dots,b^m)\in B$, then $(x^1,\dots,x^n,b^1,\dots,b^m)\in\{x\}\times B$, and so there is an $i\in I$ such that $(x^1,\dots,x^n,b^1,\dots,b^m)\in A_i$, hence $(b^1,\dots,b^m)\in\alpha_i$; this shows that $\{\alpha_i\}_{i\in I}$ is a cover of $B$. To prove that it is an open cover, note that if $i\in I$ and $(y^1,\dots,y^m)\in \alpha_i$, then $(x^1,\dots,x^n,y_1,\dots,y^m)\in A_i$. Since $A_i$ is open, there is a $\delta>0$ such that $$ (x^1-\delta,x^1+\delta)\times\dots\times(x^n-\delta,x^n+\delta)\times(y^1-\delta,y^1+\delta)\times\dots\times(y^m-\delta,y^m+\delta)\subset A_i \, . $$Hence, $$ (y^1-\delta,y^1+\delta)\times\dots\times(y^m-\delta,y^m+\delta)\subset\alpha_i \, , $$ and $\alpha_i$ is open.

Since $\{\alpha_i\}_{i\in I}$ is an open cover of $B$, it has a finite subcover $\{\alpha_i\}_{i\in I'}$. Then, $\{A_i\}_{i\in I'}$ is a finite subcover of $\{A_i\}_{i\in I}$, for if $(x^1,\dots,x^n,b^1,\dots,b^m)\in\{x\}\times B$, then $(b^1,\dots,b^m)\in B$, so there is an $i\in I'$ such that $(b^1,\dots,b^m)\in\alpha_i$, and consequently $(x^1,\dots,x^n,b^1,\dots,b^m)\in A_i$. This completes the proof.

Answer 1

I'd use the theorem that continuous images of compact are compact, and prove that $b\mapsto (x,b)$ is continuous. Given any neighborhood of $(x,b)$, there's a basic open neighborhood $U\times V$, whose inverse image is...

Edit: just saw the comment that continuity is disallowed. In that case, I'd still argue to convert open sets in the product to basic open sets, then take the open projection in the compact $B$, from which you get a finite subcover that lifts to a finite subcover of the original.

Answer 2

Generally speaking, if $\,A\,$ is a compact subset of a topological space $(X,\tau)$ and if $\,B\,$ is a compact subset of a topological space $(Y,\sigma)$, then the subset $\,A\times B\,$ of the topological space $\,X\times Y\,$ with respect to the product topology $\,\tau\times\sigma\,$ is compact too.

Since any singleton $\{x\}$ is a compact subset of a topological space, the theorem

If $B\subset\mathbb R^m$ is compact, and $x\in\mathbb R^n$, then it is easy to see that $\{x\}\times B\subset\mathbb R^{n+m}$ is compact.

is just a direct consequence of the compactness of the cartesian product of compact sets.

Answer 3

I don't think you need to go into that much technical detail. After all, the proof still works if $B$ is a general compact topological space (not necessarily equipped with the subspace topology from $\mathbb R^m$).

Here is how I would write it.

That $B$ is compact means that every open cover $\{V_i\}_{i\in I}$ of $B$ has a finite subcover $\{V_j\}_{j\in J}$, which means that $J$ is a finite subset of $I$ and that $\bigcup_{j\in J}V_j=B$.

To show that $\{x\}\times B$ is compact, we take ourselves some open cover $\{U_i\}_{i\in I}$ of $\{x\}\times B$. Notice that every $U_i$ is of the form $\{x\}\times V_i$ for some (possibly empty) open $V_i\subseteq B$.

As $\{V_i\}_{i\in I}$ is then an open cover of the compact $B$, it has a finite subcover $\{V_j\}_{j\in J}$. Then $\{\{x\}\times V_j\}_{j\in J}$ is an open subcover of $\{U_i\}_{i\in I}$.

We conclude that $\{x\}\times B$ is compact.

Answer 4

$\newcommand{\Reals}{\mathbf{R}}$This may be the essential proof in the original post, but to me the notation is clouding the topological idea. (I am not seeing the topological idea fully expressed in the existing answers, and conceivably the remarks below are useful to posterity.)

Throughout, let's agree that $U$ (with an index) connotes an open set in $\Reals^{n}$, $V$ connotes an open set in $\Reals^{m}$, and $O$ connotes an open set in $\Reals^{m+n}$ identified with $\Reals^{n} \times \Reals^{m}$.

If $K$ is a set, let's agree an open-covering (of $K$) is a collection $\{O_{\alpha}\}$ of open sets whose union contains $K$. (Pointedly inventing a compound noun to avoid writing "open covering" or "covering by open sets.")

Definition: Let $K$ be a set and $\{O_{\alpha}\}$ an open-covering of $K$. Let's say a finite refinement is a finite open-covering $\{O_{\beta}'\}$ of $K$ such that each $O_{\beta}'$ is contained in some set $O_{\alpha}$.

(Idea: A finite refinement is a finite open-covering obtained from the original collection by possibly shrinking sets. To my knowledge this term is not in use, though it's naturally motivated by the definition of paracompactness.)

Lemma 1: A set $K$ is compact if and only if every open-covering of $K$ has a finite refinement.

Proof: An open-covering refines itself, so if $K$ is compact then every open-covering of $K$ has a finite refinement.

Conversely, assume every open-covering of $K$ has a finite refinement, and let $\{O_{\alpha}\}$ be an arbitrary open-covering of $K$. Choose a finite refinement $\{O_{\beta}'\}$. For each index $\beta$, pick an index $\alpha := f(\beta)$ such that $O_{\beta}' \subset O_{\alpha} = O_{f(\beta)}$, and observe that $\{O_{f(\beta)}\}$ is an open-subcover of $K$. That is, every open-cover of $K$ has a finite subcover, i.e., $K$ is compact.

Now we're ready for the main part of the argument:

Lemma 2: If $B$ is a compact subset of $\Reals^{m}$, then every open-cover of $\{x\} \times B$ has a finite refinement.

Proof: Assume $\{O_{\alpha}\}$ is an open-cover of $K := \{x\} \times B$. For each point $(x, y)$ in $K$, there is an index $\alpha$ such that $(x, y) \in O_{\alpha}$. Because product open sets are a basis for the ambient topology, there exist open sets $U_{\alpha}$ and $V_{\alpha}$ such that $$ (x, y) \in U_{\alpha} \times V_{\alpha} \subseteq O_{\alpha}. $$ The collection $\{V_{\alpha}\}$ is an open-cover of $B$, so it has a finite subcover by compactness of $B$. The intersection $U$ of the corresponding sets $U_{\alpha}$ is open as a finite intersection of open sets, so the corresponding products $U \times V_{\alpha}$ form a finite refinement of $\{O_{\alpha}\}$.

It now follows from Lemma 1 that $K$ itself is compact.

Answer 5

Let $U_\alpha$ be an open cover of $\{x\} \times B$. Let $P_\alpha = \{ b \in B \mid (x, b) \in U_\alpha \}$.

Since $U_\alpha$ is open, it is straightforward to check that the $P_\alpha$ are open. (If $(x,b) \in U_\alpha$, then there is some $\epsilon >0$ such that $(x,b) \subset \{x\} \times B(b,\epsilon) \subset U_\alpha$, and so $B(b,\epsilon) \subset P_\alpha$.)

Hence there is a finite subcover $P_{\alpha_1},...,P_{\alpha_m}$ of $B$ and so $U_{\alpha_1},...,U_{\alpha_m}$ is a finite subcover of $\{x\} \times B$.

Answer 6

Here is a short, conceptual proof, based on principles of topology. Perhaps something like this is what Spivak had in mind.

Step 1: The function $f : B \to \{x\} \times B$, given by $f(b)=(x,b)$, is a homeomorphism.

Proof: $f$ is obviously a bijection. Also, from the definition of the product topology, $U \subset B$ is open if and only if $f(U) = \{x\} \times U \subset \{x\} \times B$ is open.

Step 2: For every homeomorphism $f : X \to Y$, $X$ is compact if and only if $Y$ is compact.

Proof: For the "only if" direction, assume $X$ is compact. To prove $Y$ is compact, let $\{U_i\}_{i \in I}$ be an open covering of $Y$. Since $f$ is continuous, it follows that $\{f^{-1}(U_i)\}_{i \in I}$ is an open covering of $X$. Since $X$ is compact, some finite subset $\{f^{-1}(U_{i_1}),..., f^{-1}(U_{i_k})\}$ covers $X$. Since $f^{-1}$ is continuous, and since $f(f^{-1})(A)=A$ for any subset $A \subset Y$, it follows that $\{U_{i_1},\ldots,U_{i_k}\}$ is a finite subset of the original covering which still covers $Y$.

For the converse "if" direction, apply the previous argument to the inverse function $f^{-1} : Y \to X$.