Suppose we have two metric spaces and , and we have a function which is continuous on . If is a connected subset of , then it's easy enough to show that is connected.
However, suppose there is a function which is only continuous on , but not necessarily the rest of . How do we show that is connected?
Any help would be much appreciated. Thanks!
So does anyone else have an idea?
I was thinking that maybe we could show that for any function , if the restriction of to is continuous, then there is a function which is continuous on , and whose restriction to is equal to that of . In that case, I could show that is connected implies is connected. However, I do not know how to prove that the continuous restriction of a function to may be expanded to a function continuous on . Heck, I'm not even sure it's true!
Help would be much appreciated!
I don't know what definition of connectedness you're using.... But a topological space is connected if it can not be written as a disjoint union of 2 non-empty open sets U,V. Thus not connected means we can find such U,V.If is disconnected, then we cannot assume that for disjoint open sets and
It seems like you haven't yet fully understood the definition of connectedness, but there is another equivalent definition which can be helpful here:
A set in a top. space is disconnected iff there exists a continuous function from onto , where this last space is the
usual two-points one with the inherited topology from the usual topology in .
Let us go back now to your problem: if is disconnected there exists continous and onto, but then
is cont. and onto, contradiction.
I understand the definition of connectedness which I related earlier. Although I haven't gone and proved it, I can see how your alternate definition might be equivalent.
That said, I figured out how to prove the contrapositive: If is disconnected, where is continuous on , then there are disjoint open sets covering with and nonempty. Putting and satisfies disconnectedness for .
Thanks for the help, guys!