# What am I missing? (continuity)

• Jul 21st 2013, 01:54 PM
krcmd
What am I missing? (continuity)
Munkres "Analysis on Manifolds "P. 30 #5 asks: let f: X- > Y. Show that f is continuous if and only if for each x [in] X there is a neighborhood U of x such that flU is continuous. I can understand I think how to prove the condition is necessary. But since x is in U doesn't the condition state "f is continuous for all x in X?" (so why need to prove the condition is sufficient?)
• Jul 21st 2013, 11:03 PM
chiro
Re: What am I missing? (continuity)
Hey krcmd.

What do you mean by f|U? Is this f \ U?
• Jul 22nd 2013, 12:50 AM
topsquark
Re: What am I missing? (continuity)
I don't know the notation well but it seems that f|U should be f: U --> Y: f(U)? Is this the meaning for $f|_U$?

-Dan
• Jul 22nd 2013, 12:26 PM
HallsofIvy
Re: What am I missing? (continuity)
Yes, that is how I would interpret f|U: "f restricted to U".

krcmd, the statement "f is continuous (on X) if and only if for each x [in] X there is a neighborhood U of x such that flU is continuous"
has, of course, two parts:
1) If, for each x in X, there is a neighborhood U of x such that f|U is continuous then f is continuous in X".
That is the "meat" of the theorem. And
2) If f is continuous in X, then there is a neighborhood U of x such that f|U is continuous".
Yes, this follows immediately from the fact that f is continuous on X. There is a technical detail to deal with: strictly speaking, "f|U" is a different function from f.
• Jul 22nd 2013, 02:13 PM
krcmd
Re: What am I missing? (continuity)
Thank you.
• Jul 23rd 2013, 04:18 PM
krcmd
Re: What am I missing? (continuity)
How is this reasoning? Form the union of all the open sets U within which f is continuous; the union U' is also open. X-U' consists of the boundary and the exterior of U'. Neither the boundary nor the exterior have any points with a neighborhood U where f is continuous. So X-U' is null, and X=U'

I still have the nagging feeling that I'm missing something very important here.