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

1. ## 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!

2. Can you show that $\{x\in D:f(x)\le 1\}$ is a closed set?

3. how? can we say that because it is a uniformly continuous function?

4. So if $0 \leq f(x_n) \leq 1$ with $x_n \rightarrow x$, with f continuous, then would we have $0 \leq f(x) \leq 1$?

5. 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!
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\to \mathbb{R}" alt="f\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.

6. [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

7. Originally Posted by Drexel28
I think the point Plato is trying to make is that if $f\to \mathbb{R}" alt="f\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.
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.

8. Originally Posted by Drexel28
Startling resemblance.

9. 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

10. 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"

11. 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

12. 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.