At Showing the "traditional" Hilbert Cube $C$ is homeomorphic to $[-1,1]^{\mathbb{N}}$, the questioner states "$[-1,1]^{\mathbb{N}}$ ... is, by Tychonoff, compact when endowed with the metric $d(x,y)=\sum_{j \in \mathbb{N}}\frac{1}{2^j}\left\lvert x_j-y_j\right\rvert$". How is that proved? Presumably one would need to show that the topology induced by this given metric is equivalent to the product topology in order for Tychonoff's theorem to apply. Beyond that it's rather unclear to me.
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You are correct: their point is that the metric and product topologies on $[-1,1]^{\Bbb N}$ are the same. And therefore by Tychonoff, that topology is compact.
Let $x \in U$, an open set in the product topology. $\pi_j(U) = [-1,1]$ except for a finite number of indexes $j$. For each of those, $\pi_j(U)$ is an open neighborhood of $x_j$. You can choose $r_j > 0$ small enough that $(x_j -r_j, x_j+r_j) \subseteq \pi_j(U)$. And since there only finitely many, you can choose $r > 0$ such that $r < 2^{-j}r_j$ for each $j$. It follows that the ball $B(x, r) \subseteq U$. Since this is true of all such $x \in U, U$ is open in the metric topology.
Conversely, let $V$ be open in the metric topology, and let $x \in V$. There is some $r$ such that $B(x,r) \subseteq V$. And there is an $N$ such that $\sum_{j\ge N}\frac 2{2^j} < \frac r2$. If we choose $$U = \left(\prod_{j< N}\left(x_i-\frac r2, x_i+\frac r2\right)\cap[-1,1]\right)\times \prod_{j\ge N} [-1,1]$$ Hence $U \subseteq B(x,r) \subseteq V$. But $U$ is an open neighborhood of $x$ in the product topology. Since this can be accomplished for all $x \in V, V$ is open in the product topology.
Therefore the two topologies are in fact the same.
Paul Sinclair
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