In Urysohn's Metrization Theorem we try to show that a space $X$ is metrizable by constructing some function $F: X\rightarrow H$ which is an embedding. Metrizability would then follow from the fact that $X$ is homeomorphic to a subspace of the Hilbert Cube.

Beneath, they try to show that $F$ is an embedding by showing that it is one-to-one, continuous andan open function.

Isn't it sufficient to show that $F$ is one-to-one and continuous? Why do we also need to show that $F$ is an open mapping? Is this the case for any embedding or is it just the case within this Theorem?

Here is the proof up until the point of constructing $F$.

Let $X$ be a second countable regular space with countable basis $\mathcal{B}=\{B_n\}_{n=1}^{\infty}$. We can easily prove $X$ is normal. Consider the collection of all ordered pairs $(i,j)$ of integers for which $\bar{B_i}\subset B_j$. By Urysohn's Lemma, there is for each such pair $(i,j)$ a Urysohn function $f: X \rightarrow [0,1]$ such that $f(\bar{B_i})=0$, $ f(X\setminus B_j)=1$.

Let $\mathcal{F}$ denote such a collection of Urysohn functions having one member for each ordered pair $(i,j)$ for which $\bar{B_i}\subset B_j$. Since $\mathcal{F}$ is countable, then it can be indexed by the set of positive integers.

Define a function $F: X \rightarrow H$ from $X$ into Hilbert space $H$ by
$$ F(x) = \left( f_1(x), \frac{f_2(x)}{2}, \frac{f_3(x)}{3}, \dots\right), x \in X$$

Thus the coordinates of $F(x)$ are determined by the values of the members of $\mathcal{F}$ at $x$; each value $f_n(x)$ is divided by $n$ to insure that $F(x)$ is a member of $H$:

$$ \sum_{n=1}^{\infty}\left( \frac{f_n(x)}{n} \right)^2 \leq \sum_{n=1}^{\infty} \frac{1}{n^2}$$

so the sum of the squares of the coordinates of $F(x)$ is a convergent series of real numbers.

To show that $F$ is an embedding, it is sufficient to show that $F$ is a one-to-one, continuous and open function to the subspace $F(X)$ of $H$. Then the metrizability of $X$ will follow from the fact that $X$ is homeomorphic to a subspace of the metric space $H$.

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    $\begingroup$ To be an embedding $F$ should be an homeomorphism onto its image so it should have a continuous inverse. $\endgroup$ – Sigur Feb 4 '13 at 20:47

A map $f:X\to Y$ that is one-to-one, continuous, and open is a homeomorphism of $X$ onto $f[X]$; that’s the definition of an embedding of $X$ into $Y$. To see why it’s not enough to say that $f$ is one-to-one and continuous, let $X$ be the reals with the discrete topology; then $f:X\to\Bbb R:x\mapsto x$ is one-to-one, continuous, and onto, but it’s certainly not an embedding, because $f[X]$ is not homeomorphic to $X$.

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  • $\begingroup$ So it is the same as to say that $F$ is a continuous bijection? An open mapping corresponds with an continuous inverse function if I'm not wrong. $\endgroup$ – onimoni Feb 4 '13 at 22:02
  • $\begingroup$ @omar: The $f$ in my example is a continuous bijection that is not open. But yes, if you have a continuous bijection, it’s open if and only if its inverse is continuous, i.e., if and only if it’s a homeomorphism. $\endgroup$ – Brian M. Scott Feb 4 '13 at 22:05

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