I want to build an intuition about Tychonoff, and for that it would be great to have an intuitive proof of why the finite product of compact spaces is compact.
Does anyone here have a short, intuitive proof? (extra points for using pictures!)
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In this answer I prove the theorem by Kuratowski that $X$ is compact iff for all spaces $Y$, the projection $\pi_Y : X \times Y \to Y$ is a closed map.
This has finite Tikhonov/Tychonoff as an easy consequence: suppose $X,Y$ are compact, and let $Z$ be any space. Then $\pi_Z: (X \times Y) \times Z \to Z$ is the composition of two closed maps $\pi_2: X \times (Y \times Z) \to Y \times Z$ and $\pi_2: Y \times Z \to Z$, making the obvious homeomorphic identifications like $(X \times Y) \times Z \simeq X \times (Y \times Z)$ etc. We use the compactness of $X$ and $Y$ to see that both $\pi_2$ are closed. And the composition of closed maps is closed. Then $\pi_Z: (X \times Y) \times Z \to Z$ is closed and the reverse then shows (as this holds for all $Z$) that $X \times Y$ is compact.
Henno Brandsma
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If you take "every net has a convergent subnet" definition of compactness, then it is straightforward: a take any net $(x_{1,i},x_{2,i},x_{3,i},\ldots,x_{n,i})_{i\in I}$ in $X_1\times \ldots \times X_n$, and then take successive subnets converging in the first, the first two, the first three, ... , the first $n$ coordinates. A net convergent in all coordinates is simply convergent, and we are done.
tomasz
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What is a net? and a convergent subnet? – Jsevillamol May 16 '17 at 13:51
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1A net is the same as a sequence, only it's indexed by an arbitrary partially ordered, directed set. A subnet of a net $(x_i){i\in I}$ is a net $(y_j){j\in J}$ such that there is a function $f\colon J\to I$ which is monotone and cofinal (very roughly speaking, you pick only some terms of $(x_i)i$, perhaps choosing some of them several times over, and perhaps drop some relations between their indices). A net $(x_i){i\in I}$ is convergent to $x$ is for every neighbourhood $U$ of $x$, the set of $i\in I$ for which $x_i\in U$ contains a final segment of $I$. – tomasz May 16 '17 at 13:53
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1@Jsevillamol: See https://en.wikipedia.org/wiki/Net_(mathematics) – tomasz May 16 '17 at 13:55
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We only need to consider the case where we have two compact spaces; the finite case then follows by induction. The proof is mostly good bookkeeping:
Let $X$ and $Y$ be compact spaces. Let $\mathcal{O} = \{O_i \mid i \in I\}$ be an open cover of $X \times Y$ in the product topology (generated by all sets of the form $U \times V$, where $U \subseteq X, V \subseteq Y$ are open).
Consider $x \in X$ fixed for now. Then for each $y \in Y$ we pick $O_{i(x,y)} \in \mathcal{O}$ such that $(x,y) \in O_{i(x,y)}$. As the product topology is generated by open squares, there are open $U_{i(x,y)} \subseteq X$ and $V_{i(x,y)} \subseteq Y$ such that $$(x,y) \in U_{i(x,y)} \times V_{i(x,y)} \subseteq O_{i(x,y)}\text{.}$$
As the $V_{i(x,y)}$, ($y \in Y$) (for this fixed $x$) form an open cover of the compact space $Y$, there is a finite subset $F_x \subseteq Y$ such that
$$Y = \cup \{V_{i(x,y)} \mid y \in F_x\}$$ and we define $$U_x = \cap \{U_{i(x,y)}\mid y \in F_x\}\text{.}$$
Every $U_x$ is open as a finite intersection of open sets. This we can do for any $x \in X$ and this defines an open cover for $X$. So finitely many cover $X$ by compactness and we have a finite set $F \subseteq X$ such that $$X = \cup \{U_x: x \in F\}\text{.}$$
I now claim that
$$\mathcal{O}'= \{O_{i(x,y)}\mid x \in F, y \in F_x\}$$ is the required finite subcover for $\mathcal{O}$. Finiteness is clear, now check it's a cover:
Let $(x,y) \in X \times Y$, then there is some $p \in F$ with $x \in U_p$. Then $y \in V_{i(p,q)}$ for some $q \in F_p$. As $U_p \subseteq U_{i(p,q)}$ by construction (as intersection) we have $(x,y) \in U_{i(p,q)} \times V_{i(p,q)} \subseteq O_{i(p,q)}$ and as $p \in F, q \in F_p$ this set $O_{i(p,q)}$ is in $\mathcal{O}'$ as required.
So $X \times Y$ is compact.
Henno Brandsma
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I have been tinkering around this proof to see if I can simplify it somehow. How about we forget entirely about the square open sets and just focus on the projections as open applications? The proof would go then like this: for each $x\in X$, there is a finite subcover $F_x\subset O$ of the compact subset ${x}\times Y$. Now, the set ${ \cap \pi_X F_x : x \in X}$ is an open cover of $X$, and thus there exists a finite subcover $F$ which we will label by the $x$'s which generate the open set. Now $\cup_{x\in F} F_x$ is a finite subcover of $X\times Y$. Does this convince you? – Jsevillamol May 16 '17 at 13:13
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@Jsevillamol Can you show that your $\cup_{x \in F} F_x$ covers $X \times Y$? If so, in detail? $(p,q)$ any point, then $p \in U_{x}$ for some $x \in F$. And then? – Henno Brandsma May 16 '17 at 18:07
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Let $(p,q)$ be any point. Then there is an $x\in F$ such that $p\in \cap \pi_X F_x$, and thus $p \in \pi_X A$ for every $A \in F_x$. On the other hand, $F_x$ covers $Y$ and therefore there is a set $A\in F_x$ such that $(p,q)\in A$. – Jsevillamol May 16 '17 at 18:20
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1@Jsevillamol $F_x$ covers ${x} \times Y$ not $Y$, and $(p,q) \notin {x} \times Y$. – Henno Brandsma May 16 '17 at 18:26
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Dang it you are right. This shows that there is a $(p,y)$ in $A$ and a $(x,q)$ in $A$, but they do not need to coincide. – Jsevillamol May 16 '17 at 20:41
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In fact this reveals why the square open sets are needed: if we can have $A$ be square, then if A has $(p,y)$ and $(x,q)$ for some $x,y$ it must also contain $(p,q)$. – Jsevillamol May 16 '17 at 20:45
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