# Prove the image of a connected set is connected.

• Apr 6th 2007, 01:02 PM
Prove the image of a connected set is connected.
Problem: Let f:R^n -> R^m be continuous. Suppose A is a connected subset of R^n. Prove that f(A) is connected.

My proof:

Now f(A) = {f(x) : x in A}, and assume f to be an increasing function.

Now assume to the contrary that f(A) is not connected.
Let the two points d and e be in the set of f(A), then there exists open sets U and V in R^n such that they separate d and e.

Now d = f(a) and e = f(b) for a and b in A. Now assume a < b.

Since A is connected, A has Intermedian Value Property, that is, for every continuous function f, there exists a point z in A such that f(a) <= f(z) <= f(b) by the Intermedian Value Theorem.

But that contradict our assumption as it would be impossible for U and V to separate f(A).

Therefore we conclude f(A) is connected.

Q.E.D.

Yeah, I know this one is not really proper... But this is the best I could sequeeze out after almost two days of thinking...

KK
• Apr 6th 2007, 01:16 PM
CaptainBlack
Quote:

Problem: Let f:R^n -> R^m be continuous. Suppose A is a connected subset of R^n. Prove that f(A) is connected.

My proof:

Now f(A) = {f(x) : x in A}, and assume f to be an increasing function.

Now assume to the contrary that f(A) is not connected.
Let the two points d and e be in the set of f(A), then there exists open sets U and V in R^n such that they separate d and e.

Now d = f(a) and e = f(b) for a and b in A. Now assume a < b.

Since A is connected, A has Intermedian Value Property, that is, for every continuous function f, there exists a point z in A such that f(a) <= f(z) <= f(b) by the Intermedian Value Theorem.

But that contradict our assumption as it would be impossible for U and V to separate f(A).

Therefore we conclude f(A) is connected.

Q.E.D.

Yeah, I know this one is not really proper... But this is the best I could sequeeze out after almost two days of thinking...

KK

Sketch of proof:

Suppose otherwise. The let a and b be points in disjoint components of f(A).

Let C be a curve connecting a to b in A, then f(C) is in f(A), but now f(C)
is a curve in R^m conecting f(a) to f(b), but this contradicts the assumption
that a and b are in disconnected components of f(A).

(here the notion of curve implies continuity, and the continuity of f ensures
that f(C) is a curve in the appropriate sense)

RonL
• Apr 6th 2007, 03:19 PM
Plato
Once again, not knowing what theorems you know it is very hard to give you a proof. In the attached pdf file I have given the standard proof that will work in any topological space. Again this may be of any use to you if you do not have the theorems used.
• Apr 6th 2007, 05:45 PM
So far, our class has learned:

Defintion:

A set B not connected if there are open sets U and V such that:

B interset U is not empty, B interset V is not empty.
(B interset) U U (B interset V) equals to B.
(B interset U) interset (B interset V) is empty.

Theorems:

1. A is connected iff A has Intermedian Value Property.
2. A is connected if A is pathwise connected.
3. A is pathwise connected iff A is an interval.
4. Let ACR^n be pathwise connected, let f:A->R^n be continuous, then f(A) is PC.
5. Every convex set is pathwise connected.

That is all we have learn so far. Would that help?
• Apr 7th 2007, 03:52 AM
Plato
Quote: