# compact subset of R

• Mar 3rd 2009, 01:55 PM
junipers232
compact subset of R
Suppose that $\displaystyle A$ is a compact subset of $\displaystyle \mathbb{R}$ and $\displaystyle f: A \rightarrow \mathbb{R}$ is continuous. Prove that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is a compact subset of $\displaystyle \mathbb{R}$.

I know that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is bounded because $\displaystyle A$ is bounded, so I guess I just need to prove that $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is closed. I was thinking to take a sequence in the set and show that it converges in the set. This is what is confusing me because I don't know how to do this part. Thanks for any help in advance.
• Mar 3rd 2009, 02:45 PM
Plato
Use the notation: $\displaystyle B = \left\{ {x \in A:0 \leqslant f(x) \leqslant 1} \right\}$.
Theorem: Any closed subset of a compact set is compact.
You should have proved that theorem, if not it is easily proven.
Thus, you want to prove that the set $\displaystyle B$ is closed.
If $\displaystyle B$ were not closed then some limit point, $\displaystyle x_0$, of $\displaystyle B$ and $\displaystyle x_0 \notin B$.
Then there is a sequence of distinct points from $\displaystyle B$ which converges to $\displaystyle x_0$.
Remembering that $\displaystyle f$ is continuous, there is a contradiction there.
• Mar 3rd 2009, 02:56 PM
aliceinwonderland
Quote:

Originally Posted by junipers232
Suppose that $\displaystyle A$ is a compact subset of $\displaystyle \mathbb{R}$ and $\displaystyle f: A \rightarrow \mathbb{R}$ is continuous. Prove that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is a compact subset of $\displaystyle \mathbb{R}$.

I know that the set $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is bounded because $\displaystyle A$ is bounded, so I guess I just need to prove that $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is closed. I was thinking to take a sequence in the set and show that it converges in the set. This is what is confusing me because I don't know how to do this part. Thanks for any help in advance.

Lemma 1. Each closed subset of a compact space is compact.

Since f is continuous, each closed subset C of $\displaystyle \mathbb{R}$, $\displaystyle f^{-1}(C)$ is closed in A.

By lemma 1, $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is compact. Now, you'll see that $\displaystyle \{x \in A | 0 \leq f(x) \leq 1 \}$ is a compact subset of $\displaystyle \mathbb{R}$.