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In Urysohn's Metrization Theorem, at some point we define a function $F : X \rightarrow H$ from the space X into Hilbert space $H$. Whereupon we need to show that $F$ is an embedding. To show this, it apparently suffices to show that $F$ is

  • one-to-one
  • continuous
  • an open mapping.

I don't see how the last two differ?

Continuity is proven with open sets, so is an open mapping?

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up vote 3 down vote accepted

A map is continuous if preimage of open sets are open sets and it is open if direct image of open set are open sets.

More precisely, $f:X\to Y$ is continuous if $f^{-1}(V)\subset X$ is open for any open set $V\subset Y$.

It is open if $f(U)\subset Y$ is open for any open set $U\subset X$.

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You show it is an open map to apply this to its inverse. A continuous, bijective map which is open has a continuous inverse (just look at the definition), and is so a homeomorphism.

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An open mapping is a function $f:X\to Y$ such that if $U\subseteq X$ is open, then $f(U)$ is open. A continuous function is one such that if $V\subseteq Y$ is open, then $f^{-1}(V)$ is open.

For an example of a continuous function that is not open, take $f:\mathbb R\to \mathbb R$ defined by $f(x)=0$. For an example of an open function that is not continuous, give $X$ the trivial topology and let $Y$ be the same set with the discrete topology. If $|X|>1$ then the identity map $f:X\to Y$ is open but not continuous. For an example of such a function on $\mathbb R$ with the usual topology, see here.

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