Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

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 $U\subset X$.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.