Math Is Fun Forum

AlexGo

Guest

Query with proof

Claim: A continuous function on a closed, bounded interval is bounded.

Proof: Let f: [a,b] --> R. Suppose the statement is false, i.e. that for every positive integer n, there is x_n ∈ [a,b] such that |f(x_n)| > n.

By Bolzano-Weierstrass, x_n has a convergent subsequence x_(n_j) --> x, a ≤ x ≤ b.

Now |f(x_(n_j)| > n_j, and since f is continuous, f(x_(n_j)) --> f(x). But |f(x_(n_j)| > n_j --> ∞. Contradiction.

My problem

What I'm unsure about is what exactly is the contradiction? How do we know that f(x) isn't infinite? It seems "obvious" that it can't be, as f is continuous, but I've never seen a proof of this. Also, we are already assuming f isn't bounded... it all seems slightly circular to me. What am I missing?

Thanks.

AlexGo

Guest

Re: Query with proof

Duh. f(x_(n_j)) C-O-N-V-E-R-G-E-S to f(x). So of course f(x) is finite. I'm an idiot.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Query with proof

It's actually pretty easy to prove the generalization: If f is a continuous map from a compact space X, then f(X) is compact.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."