# Why is this embedding a homeomorphism?

In Robert Israel's answer, it's posited that the natural embedding $\iota$ of $B = \{y \in \ell_\infty: \|y\| \le 1\}$ into $P = \prod_{x \in \ell_1} [-\|x\|,\|x\|]$ is a homeomorphism. Coming back to it a few days later, I realize I don't quite see immediately why this is. Is there a more explicit reason that such a map is a homeomorphism?

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Reading the earlier question, I think Prof. Israel means that the function $\iota:B\to P$ is an embedding, in other words $\iota:B\to\iota[B]$ is a homeomorphism.

That $\iota:B\to P$ is an embedding is a particular case of this general theorem:

Let $X$ be a set, $Y_i,i\in I$ topological spaces and $f_i:X\to Y_i,i\in I$ maps. Assume that for every $x_1,x_2\in X,x_1\neq x_2$ there is $i\in I$ such that $f_i(x_1)\neq f_i(x_2)$ (i.e. $\{f_i:i\in I\}$ separates points). Then the function $F:X\to\prod_{i\in I}Y_i$ defined by $F(x)_i=f_i(x)$ is an embedding where $X$ is equipped with the coarsest topology which makes all the $f_i$'s continuous.

The book General Topology by Stephen Willard has a proof of this fact (Theorem 8.12, page 56).

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Thanks LostInMath. So why does the fact that $\iota$ is an embedding immediately imply that $B\mapsto\iota[B]$ is a homeomorphism? Is this a well known fact? – Gotye Sep 30 '11 at 22:36
@Gotye: An embedding of $X$ into $Y$ is by definition a homeomorphism of $X$ onto a subspace of $Y$. – Brian M. Scott Sep 30 '11 at 22:40
@BrianM.Scott, I see, I thought it was an injective map with no other properties. – Gotye Sep 30 '11 at 22:44
@Gotye: I just wanted to add that the theorem in question has some nice applications. An important one is Urysohn-Tychonoff metrization theorem. Another one which is quite surprising at first is that every topological space can be embedded in a power of a certain three-point topological space $Y$, in other words for every topological space $X$ there is a set $S$ and an embedding $X\to Y^S$. So topology, as a branch of mathematics, is nothing else but the study of the subspaces of the powers of a certain simple finite space $Y$. That must be quite trivial, mustn't it ;). – LostInMath Sep 30 '11 at 23:16