# If a unit ball is compact then why a ball of radius 5 has to be compact too?

So if I use the definition of compactness that every open cover has a finite sub-cover, then as the unit ball is compact , there exists a finite subcover. But if I increase the radius of the ball, why does it still need to be compact. Intuitively speaking can't I just take very small sized and large number open sets in such a way that there is no finite sub-cover. I know that the answer to this question is no but I don't see why?

Can someone please throw some light?

• This is not true in general metric spaces. Are you talking about Euclidian vector spaces ($\mathbb R^n$ with the standard metric, for example)? – sranthrop Jul 5 '15 at 22:51
• @sranthrop My question was pertaining to Banach Spaces – user3503589 Jul 5 '15 at 22:53
• Note that in infinite-dimensional Banach spaces (which are the most interesting ones), the closed unit ball is not compact, nor is any other ball. – Nate Eldredge Jul 6 '15 at 7:29
• @NateEldredge Thanks , I do know that a closed unit ball is compact iff the space is finite dimensional – user3503589 Jul 6 '15 at 7:32

Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1. Therefore it has a finite subcover. Now reflate the subcover by a factor of 5. Et Voila! It covers the ball of radius 5.

• Strange, and seemingly unwarranted downvote on three posts here. No comment either. – ncmathsadist Jul 8 '15 at 0:13

Strictly positive scaling is a homeomorphism on closed balls. So either none is compact or all.

• A homeomorphism is a continuous bijection with a continuous inverse . What exactly do you mean by ""Strictly positive scaling is a homeomorphism on closed balls."" – user3503589 Jul 6 '15 at 19:17
• I mean that for scaling by any nontrivial scalar defines a homeomorphism; that is continuous with continuous inverse as you mentioned. Scaling by zero is only continuous. So one can conclude that for compact unit balls the origin itself is compact. So it is always compact as singleton. – C-Star-W-Star Jul 7 '15 at 0:01

If $B_5(x_0)$ is a closed ball with center $x_0$ and radius 5 in a Banach space, then $f(x):=5x+x_0$ maps the unit ball continuously onto $B_5(x_0)$. If the unit ball is compact, then so is $B_5(x_0)$ (as the image of a compact set under a continuous function).

Given an open cover of the closed radius 5 ball, scale it down by a factor of $\frac15$. You obtain an open cover of the closed unit ball. By the compactness of that ball, you can pick a finite subcover. That finite subcover, scaled back up by a factor fo $5$, is a finite subcover of the original open cover of the radius 5 ball.

If $\mathcal U_1 = \{ U \}$ is an open cover of the unit ball, then $\mathcal U_5 = \{ 5U \}$ is an open cover of the ball of radius $5$ where $5U = \{ 5x : x \in U \}$. And vice versa in the obvious way.

Now the result follows: any open cover of the $5$-radius ball can be mapped to a cover of the unit ball, which has a finite sub-cover; then map back that finite sub-cover to a finite sub-cover of the $5$-radius ball.