# how to prove that x: 0<=f(x)<=1 is compact

• Apr 4th 2010, 04:00 PM
shuxue
how to prove that x: 0<=f(x)<=1 is compact
Suppose f: D-R is continuous with D compact. prove that x: 0<=f(x)<=1 is compact.

what i got so far is that since D is compact, then D is closed and bounded. and f is uniformly continuous and f(D) is compact. but i dont how to show that x: 0<=f(x)<=1 is compact.

Thanks!
• Apr 4th 2010, 04:23 PM
Plato
Can you show that $\{x\in D:f(x)\le 1\}$ is a closed set?
• Apr 4th 2010, 05:15 PM
shuxue
how? can we say that because it is a uniformly continuous function?
• Apr 4th 2010, 07:52 PM
So if $0 \leq f(x_n) \leq 1$ with $x_n \rightarrow x$, with f continuous, then would we have $0 \leq f(x) \leq 1$?
• Apr 4th 2010, 07:54 PM
Drexel28
Quote:

Originally Posted by shuxue
Suppose f: D-R is continuous with D compact. prove that x: 0<=f(x)<=1 is compact.

what i got so far is that since D is compact, then D is closed and bounded. and f is uniformly continuous and f(D) is compact. but i dont how to show that x: 0<=f(x)<=1 is compact.

Thanks!

Quote:

Originally Posted by Plato
Can you show that $\{x\in D:f(x)\le 1\}$ is a closed set?

I think the point Plato is trying to make is that if $f:D\to \mathbb{R}$ is continuous we have that $f^{-1}\left([0,1]\right)\subseteq D$ is closed. But, a closed subspace of a compact space is compact.
• Apr 4th 2010, 11:02 PM
FancyMouse
[0,1] is closed in R. Inverse image of closed set under continuous map is closed. Closed subset of a compact space is compact.

>then D is closed and bounded
That's already false since you don't have any relationship between D and R, I mean whether D is a subset of R or not. The Bolzano–Weierstrass theorem only applies when D is a subset of R^n
• Apr 4th 2010, 11:03 PM
Drexel28
Quote:

Originally Posted by Drexel28
I think the point Plato is trying to make is that if $f:D\to \mathbb{R}$ is continuous we have that $f^{-1}\left([0,1]\right)\subseteq D$ is closed. But, a closed subspace of a compact space is compact.

Quote:

Originally Posted by FancyMouse
[0,1] is closed in R. Inverse image of closed set under continuous map is closed. Closed subset of a compact space is compact.

Startling resemblance.
• Apr 4th 2010, 11:08 PM
FancyMouse
Quote:

Originally Posted by Drexel28
Startling resemblance.

• Apr 4th 2010, 11:29 PM
Drexel28
Quote:

Originally Posted by FancyMouse
That's already false since you don't have any relationship between D and R, I mean whether D is a subset of R or not. The Bolzano–Weierstrass theorem only applies when D is a subset of R^n

I don't follow. In any topological space we have that the inverse image of a closed set under a continuous map is closed, and that the closed subspace of a compact space is compact. There is no need for the B.W.P. here
• Apr 4th 2010, 11:37 PM
FancyMouse
Quote:

Originally Posted by Drexel28
I don't follow. In any topological space we have that the inverse image of a closed set under a continuous map is closed, and that the closed subspace of a compact space is compact. There is no need for the B.W.P. here

I'm commenting on shuxue's statement that "what i got so far is that since D is compact, then D is closed and bounded"
• Apr 6th 2010, 05:32 PM
alice8675309
Does anyone know/have the actual proof to this problem? It looks interesting and I was wondering if I could see the proof. (I'd attempt it but i don't want to start mixing my classes. Therefore i kind of just want to compare and contrast)
Thanks
• Apr 6th 2010, 05:39 PM
Drexel28
Quote:

Originally Posted by alice8675309
Does anyone know/have the actual proof to this problem? It looks interesting and I was wondering if I could see the proof. (I'd attempt it but i don't want to start mixing my classes. Therefore i kind of just want to compare and contrast)
Thanks

It's exactly what I said. Since $f$ is continuous we have that $f^{-1}([0,1])=\left\{x:0\leqslant f(x)\leqslant 1\right\}$ is a closed subspace of $D$. But, $D$ is compact and so any closed subspace is automatically compact.