# Continuity in metric spaces

• Sep 23rd 2011, 02:21 PM
Jame
Continuity in metric spaces
Let $\displaystyle X$ be a metric space, s.t. $\displaystyle X = C \cup D$ where $\displaystyle C$ and $\displaystyle D$ are closed sets.

Suppose $\displaystyle f: X -> Y$ is continuous on $\displaystyle C$ and $\displaystyle D$.

Show $\displaystyle f$ is continuous on $\displaystyle X$

My proof:

$\displaystyle x \in X$ implies $\displaystyle x \in C$, or $\displaystyle x \in D$ or $\displaystyle x \in C \cap D$

(I am just going to treat the elements in the intersection as belonging to C or D only.)

Let $\displaystyle V \subset Y$ be closed in $\displaystyle Y$.

By continuity of $\displaystyle f$:

$\displaystyle f^{-1}(V)$ is closed in $\displaystyle C$
$\displaystyle f^{-1}(V)$ is closed in $\displaystyle D$

I want to say this is enough to prove $\displaystyle f$ is continuous on $\displaystyle X$.

But I think I am lacking something. I don't really use the fact that $\displaystyle C$ and $\displaystyle D$ are closed.

This theorem could say open and I could literally do the same thing. But I am 100% sure this is not true for open sets.

I'm really not sure what to do. I don't think considering complaints here helps at all.
• Sep 23rd 2011, 02:35 PM
Drexel28
Re: Continuity in metric spaces
Quote:

Originally Posted by Jame
Let $\displaystyle X$ be a metric space, s.t. $\displaystyle X = C \cup D$ where $\displaystyle C$ and $\displaystyle D$ are closed sets.

Suppose $\displaystyle f: X -> Y$ is continuous on $\displaystyle C$ and $\displaystyle D$.

Show $\displaystyle f$ is continuous on $\displaystyle X$

My proof:

$\displaystyle x \in X$ implies $\displaystyle x \in C$, or $\displaystyle x \in D$ or $\displaystyle x \in C \cap D$

(I am just going to treat the elements in the intersection as belonging to C or D only.)

Let $\displaystyle V \subset Y$ be closed in $\displaystyle Y$.

By continuity of $\displaystyle f$:

$\displaystyle f^{-1}(V)$ is closed in $\displaystyle C$
$\displaystyle f^{-1}(V)$ is closed in $\displaystyle D$

I want to say this is enough to prove $\displaystyle f$ is continuous on $\displaystyle X$.

But I think I am lacking something. I don't really use the fact that $\displaystyle C$ and $\displaystyle D$ are closed.

This theorem could say open and I could literally do the same thing. But I am 100% sure this is not true for open sets.

I'm really not sure what to do. I don't think considering complaints here helps at all.

Well, they need to agree on $\displaystyle C\cap D$...but anwaysys, so you know that $\displaystyle f^{-1}(V)=C\cap A$ and $\displaystyle f^{-1}(V)=D\cap B$ for closes sets $\displaystyle A,B\subseteq X$ but since $\displaystyle C,D$ are closed this implies that $\displaystyle C\cap A,D\cap B$ are closed in $\displaystyle X$ and since $\displaystyle f^{-1}(V)=(C\cap A)\cup (D\cap B)$ the conclusion follows.

Also, this is true for any finitely many closed sets $\displaystyle X=C_1\cup\cdots\cup C_n$ and true for arbitrary open sets $\displaystyle \displaystyle X=\bigcup_{\alpha\in\mathcal{A}}D_\alpha$.
• Sep 23rd 2011, 03:50 PM
Jame
Re: Continuity in metric spaces
Cool. Makes sense.

I have a question. Why must we union $\displaystyle C \cap A$ and $\displaystyle D \cap B$? Is knowing $\displaystyle f^{-1}(V) = C \cap A$ enough to say $\displaystyle f$ is continuous on $\displaystyle X$? $\displaystyle C \cap A$ is a closed set in $\displaystyle X$.
• Sep 23rd 2011, 04:41 PM
Drexel28
Re: Continuity in metric spaces
Quote:

Originally Posted by Jame
Cool. Makes sense.

I have a question. Why must we union $\displaystyle C \cap A$ and $\displaystyle D \cap B$? Is knowing $\displaystyle f^{-1}(V) = C \cap A$ enough to say $\displaystyle f$ is continuous on $\displaystyle X$? $\displaystyle C \cap A$ is a closed set in $\displaystyle X$.

Because technically what is meant is that $\displaystyle (f_{\mid C})^{-1}(V)=C\cap A$ where $\displaystyle f_{\mid C}$ is the restriction of $\displaystyle f$ to $\displaystyle C$, but I suppressed that.