# Conts fncs, open sets, boundry

1. May 24, 2007

### redyelloworange

1. The problem statement, all variables and given/known data
Let U and V be open sets in Rn and let f be a one-to-one mapping from U onto V (so that there is an inverse mapping f-1). Suppose that f and f-1 are both continuous. Show that for any set S whose closure is contained in U we have f(bd(S)) = bd(f(S)).

2. Relevant equations

Open sets: every point in the set is an interior point. int(S) = S. Or S contains none of its boundry points.

bd(S) = {x in Rn | B(r,x)∩S≠ø and B(r,x)∩Sc≠ø for every r>0}

3. The attempt at a solution

ie, show that the function takes a boundry to a boundry.

Let S be a set in U. The let f(S) = T. Note that f-1(T) = S.

Also, since T = {x in Rn s.t. f(x) V} and f conts, and V open, then T is open.
The same conclusion holds for S.

Thus T and S are both open. They also have the same number of elements.

I don't know where to go from here.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. May 24, 2007

### Dick

Now state the definition of continuity in terms of balls. f(x)=y is continuous at x if for every B(y,d) there is a radius e such that f(B(x,e)) is contained in B(y,d).

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