Topology - Use Componentwise Convergence Criterion to prove closed ball closed.

[t00dles23]

Homework Statement

Let r be a positive number and define F = {u in R^n | ||u|| <= r}. Use the Componentwise Convergence Criterion to prove F is closed.

Homework Equations

The Componentwise Convergence Criterion states: If {u_k} in F converges to c, then p_i(u_k) converges to p_i(c). That is, the ith component of the sequence converges to the ith component of c.

The Attempt at a Solution

What we want to show is that if {u_k} in F converges to c, then ||c|| <= r.

It's easy to show it without the componentwise convergence criterion, as follows:
||c|| <= ||c - u_k|| + ||u_k||. Taking the limit as k goes to infinity, we get ||c|| <= 0 + ||u_k|| <= r. But I need a proof that does use the CCC.

I tried this:
Suppose {u_k} in F converges to c. Then p_i(u_k) converges to p_i(c) by the componentwise convergence criterion. So:
|p_i(u_k)| <= ||u_k|| <= r
Taking the limit as k goes to infinity, we get
|p_i(c)| <= r
(p_i(c))^2 <= r^2
\sum_{i=1}^{n} (p_{i}(c))^2 \leq \sum_{i=1}^{n} r^2
\sqrt{\sum_{i=1}^{n} (p_{i}(c))^2} \leq \sqrt{nr^2}
||c|| <= (n)^(1/2)*r

but this isn't a strong enough statement, because I need ||c|| <= r. Advice please?

 

[micromass]

If p_i(u_k) --> p_i(c) for every i, then what happens to the sequence

\sqrt{\sum_{i=1}^n{p_i(u_k)^2}}

as k goes to infinity? To which number does this sequence converge?

 

[t00dles23]

I don't think I understand your hint. I mean sure, ||u_k|| --> ||c||. But I don't know what that does for me. It seems to just point back to the proof using ||c|| and the triangle inequality, but that doesn't use the componentwise convergence criterion. Can you elaborate?

 

[t00dles23]

Nevermind. I understand now. Thank you for your hint.