Continuity in metric spaces

Let be a metric space, s.t. where and are closed sets.

Suppose is continuous on and .

Show is continuous on

My proof:

implies , or or

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

Let be closed in .

By continuity of :

is closed in

is closed in

I want to say this is enough to prove is continuous on .

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

Re: Continuity in metric spaces

Quote:

Originally Posted by

**Jame** Let

be a metric space, s.t.

where

and

are closed sets.

Suppose

is continuous on

and

.

Show

is continuous on

My proof:

implies

, or

or

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

Let

be closed in

.

By continuity of

:

is closed in

is closed in

I want to say this is enough to prove

is continuous on

.

But I think I am lacking something. I don't really use the fact that

and

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 ...but anwaysys, so you know that and for closes sets but since are closed this implies that are closed in and since the conclusion follows.

Also, this is true for any finitely many closed sets and true for arbitrary open sets .

Re: Continuity in metric spaces

Cool. Makes sense.

I have a question. Why must we union and ? Is knowing enough to say is continuous on ? is a closed set in .

Re: Continuity in metric spaces

Quote:

Originally Posted by

**Jame** Cool. Makes sense.

I have a question. Why must we union

and

? Is knowing

enough to say

is continuous on

?

is a closed set in

.

Because technically what is meant is that where is the restriction of to , but I suppressed that.