Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X = A \cup B$, where $A$ and $B$ are subspaces of $X$. let $f: X \to Y$; suppose that the restricted functions $f\upharpoonright A:A\to Y$ and $f\upharpoonright B:B\to Y$ are continuous. Show that if both $A$ and $B$ are closed in $X$, then $f$ is continuous.

How does using h and g, as arbitrary functions in the hint below work?

share|improve this question
    
What is $X$? A metric space? A subset of $\mathbb{R}^n$? –  Zev Chonoles Nov 6 '11 at 9:47
1  
If $X$ is a general topological space, $\epsilon-\delta$ arguments aren’t available: they require a metric. (They’re also unnecessarily complicated, unless the only definition of continuity that you have is for metric spaces.) –  Brian M. Scott Nov 6 '11 at 9:49
2  
Use the fact that continuity can be equivalently defined as: the inverse image of a closed set is closed. –  Henno Brandsma Nov 6 '11 at 9:58

1 Answer 1

So... $X$ and $Y$ are topological spaces and you want to show that, for any closed subset $S$ of $Y$, $R=f^{-1}(S)$ is a closed subset of $X$.

Hints: obviously, $R=(R\cap A)\cup (R\cap B)$. Furthermore, $R\cap A=g^{-1}(S)$ and $R\cap B=h^{-1}(S)$ for some suitable functions $g$ and $h$. But you will need to define precisely the functions $g$ and $h$, in particular, the spaces $g$ and $h$ are defined on. A last hint: the union of two closed subsets is a closed subset.

share|improve this answer
    
Since $A$ and $B$ are closed, wouldn't checking whether inverse image of a closed set is closed give an easier proof? (As suggested in Henno's comment above.) –  Martin Sleziak Nov 6 '11 at 10:09
    
@Martin: I agree, even if the OP has to prove that characterization of continuity from some other one first. Showing that the open sets in $A$ and $B$ can be pieced together to get an open set in $X$ takes a bit more work. –  Brian M. Scott Nov 6 '11 at 10:17
    
@Martin, obviously. Thanks, I modified my post. –  Did Nov 6 '11 at 10:20
    
So, I can then write: g^1(S) U h^-1(S) = g^-1h^-1(S) and I am trying to see how I can organize the rest –  Buddy Holly Nov 6 '11 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.