Local Formulation of Continuity

[Bashyboy]

Homework Statement

Let $X$ and $Y$ be topological spaces, and let $\{U_i\}$ be a collection of open sets in $X$. If $X = \bigcup U_i$ and $f|_{U_i}$ is continuous, then $f : X \to Y$ continuous.

Homework Equations

The Attempt at a Solution

Let $x \in X$, and let $V \subseteq Y$ be some open nbhd of $f(x)$. Then there exists an $i$ such that $x \in U_i$. Since $f_{U_i}$ is continuous, there exists a set $\mathcal{O}$ that is open in $U_i$ and contains $x$ such that $f|_{U_i}(\mathcal{O}) \subseteq V$. But as $U_i$ is open in $X$, so must $\mathcal{O}$; moreover, $f(\mathcal{O}) = f_{U_i}(\mathcal{O})$ holds since $\mathcal{O}$ is completely contained in $U_i$. Therefore, $f$ must be continuous.

How does that sound?

 

[andrewkirk]

There seem to be some steps missing. What is needed is to show that $f^{-1}(V)$ is open. There are some deductions in the above proof that may relate to that, but nothing that explicitly reaches that conclusion - ie 'closes the deal'.

 

[Bashyboy]

Well, I am working with a different formulation of continuity:

For each $x \in X$ and each neighborhood $V \subseteq Y$ of $f(x)$, there is a neighborhood $U$ of $x$ such that $f(U) \subseteq V$.

Perhaps that makes a difference.