Is this ball in C([0,1]) compact?

[andrassy]

Homework Statement

Let C([0,1]) be the metric space of continuous functions on the interval [0,1] with distance = max of x over [0,1] of |f(x)-g(x)|. Is the ball of radius 1 centered around f(x) = 0 compact?

The Attempt at a Solution

I originally thought it was but now I believe that it is not compact. I'm not sure how to prove it though. I know I can use either sequential convergence or show that it isn't totally bounded, but this is where I get stuck. I know the ball has all continuous functions s.t. |f(x)| < 1. How can I go about showing it isn't sequentially compact or that it isn't totally bounded? Anyone can put me in the right direction?

 

[adriank]

Are you sure it's the open ball, with |f| < 1? In that case, it's easily not compact (it's not closed).

The closed ball would be harder, but look up Ascoli's theorem. Is the ball equicontinuous?

 

[maze]

This is a great question!

In a metric space, compactness is equivalent to sequential compactness, so we only need to determine if every sequence of points in the ball has a convergent subsequence.

In a n-dimensional vector space, you could construct a finite sequence f1, f2, ..., fn with each fi far apart from all the others by placing each fi on a different perpendicular axis. eg: in 3d f1=(1,0,0), f2=(0,1,0), f3=(0,0,1). Leting n go to infinity, in an infinite dimensional vector space theoretically you should be able to create an infinite sequence with no convergent subsequence using a similar argument.

The problem arises that C([0,1]) does not have an inner product, so you can't construct a "orthogonal" set of vectors. However, it is a normed vector space so you can effectively do the equivalent by applying Reisz's lemma to construct a sequence of unit vectors, each one of which is at least a fixed distance from the subspace spanned by all the previous vectors.