Continuity in metric spaces

**Jame** · September 23rd 2011, 02:21 PM

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

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

Show $f$ is continuous on $X$

My proof:

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

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

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

By continuity of $f$ :

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

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

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

**Drexel28** · September 23rd 2011, 02:35 PM

Originally Posted by Jame

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

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

Show $f$ is continuous on $X$

My proof:

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

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

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

By continuity of $f$ :

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

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

But I think I am lacking something. I don't really use the fact that $C$ and $D$ 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 $C\cap D$ ...but anwaysys, so you know that $f^{-1}(V)=C\cap A$ and $f^{-1}(V)=D\cap B$ for closes sets $A,B\subseteq X$ but since $C,D$ are closed this implies that $C\cap A,D\cap B$ are closed in $X$ and since $f^{-1}(V)=(C\cap A)\cup (D\cap B)$ the conclusion follows.

Also, this is true for any finitely many closed sets $X=C_1\cup\cdots\cup C_n$ and true for arbitrary open sets $\displaystyle X=\bigcup_{\alpha\in\mathcal{A}}D_\alpha$ .

**Jame** · September 23rd 2011, 03:50 PM

Cool. Makes sense.

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

**Drexel28** · September 23rd 2011, 04:41 PM

Originally Posted by Jame

Cool. Makes sense.

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

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

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