Local Continuity and Restriction

1. Aug 27, 2011

Bacle

Hi,

Let f :X-->Y ; X,Y topological spaces is any map and {Ui: i in I} is a cover for X
so that :

f|_Ui is continuous, i.e., the restriction of f to each Ui is continuous, then:

1) If I is finite , and the {Ui} are all open (all closed) , we can show f is continuous:

taking W open in Y, f^-1(W)= \/ f^-1(W /\ Ui) = is the union of open sets in X;

each W/\Ui is open, and W/\Ui is contained in Ui.

( where \/ is the union over I ; /\ is intersection over I ); same for V closed in Y.

2) If I is infinite, the argument can break down , i.e., if {Ui } is a closed cover

for X (each Ui is closed) ; f|_Ui is continuous and I is infinite, then this result

fails. Does anyone know of an example of this last?

2. Aug 27, 2011

micromass

As an example where it fails, take any discontinuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ and consider the closed cover $(\{x\})_{x\in \mathbb{R}}$, doesn't that work?

3. Aug 27, 2011

Bacle

Micromass: but how do we then define f|_x to be continuous?

4. Aug 27, 2011

micromass

Just define $f\vert_{\{x\}}:\{x\}\rightarrow \mathbb{R}:x\rightarrow f(x)$. This is clearly continuous since $\{x\}$ is indiscrete...

5. Aug 27, 2011

Bacle

Right, good point; thanks.