Expert: Paul Klarreich - 4/29/2010

Question
Can you help me prove this problem. It's for my Basic Analysis Clas.

Let f:R->R be a continuous function and let Hbe a subset ofthe range of f such that every f(x) in H there is a neighborhood V of f(x) such that V is a subset of H. Let G ={x in R such that f(x) is in H}. Prove thatfor every x in G, there is a neighborhood U of x such that U is a subset of G.

Answer
Questioner: Alex
Country: United States
Category: Advanced Math
Private: Yes
Subject: analysis
Question: Can you help me prove this problem. It's for my Basic Analysis Clas.

Let f:R -> R be a continuous function WITH RANGE Y and
let H be a subset of Y such that for every y in H there is a neighborhood V of y such that V is a subset of H.

>> In other words, H contains only interior points of Y.

Let G ={ x in R such that f(x) is in H }.

Prove that for every x0 in G, there is a neighborhood U of x0 such that U is a subset of G.

>> I.e, if  f(x0) is in H, there is a nbhd N(x0) such that f(N(x0)) is in H.

....................................................
As I recall,  f is continuous on R if f is continuous at every x0 in R,

AND

f is continuous at x0 if whenever x is near x0, f(x) is near f(x0).

SO THEN:

given some nbhd FN of f(x0), [no matter how small, they usually say] you can find a nbhd N of x0 [by making it sufficiently small] whose image is inside N.  I.e. x in N implies f(x) in FN
..........................

OK, suppose that x0 is in G, and that EVERY nbhd of x0 contains points x, such that x is not in G.  

That means that x0 is in G, and that EVERY nbhd of x0 contains points x, such that f(x) is not in H.

I think that will do it.  Look at your f(x0) in H.  Take some nbhd FN(f(x0)) in H.

We cannot find a nbhd N in R such that f(N) <= FN, because every nbhd of x0 has points such that f(x) is not in H.

Sounds complex, but it is really your old friend, the delta-epsilon stuff.