Prove that the vector space $(C[0,1],\Vert \Vert_{\infty})$ is not finite dimensional using the concept of a closed unit ball.

Question

I need to prove

The vector space of continuous real valued functions defined on a closed interval $[0,1]$ $(C[0,1],\Vert \Vert_{\infty})$, is not finite dimensional.

Using the following theorem Let X be a normed vector space and let $S = \{x ∈ X : || x || = 1\}$ be the closed unit ball in X. If S is compact then X is finite dimensional.

My Attempt: First let $X=(C[0,1],\Vert \Vert_{\infty})$ and note that $X$ has an infinite basis (as the space $P(\mathbb{R}) \subseteq X$ and $P(\mathbb{R})$ has a infinite linearly independent basis) represented as $(b_1,b_2,b_3,...)$ linearly independent elements. For each natural number $k$, let $Y_k=span\{b_1,b_2,b_3,...,b_k\}$ and let $Y_0=\{0\}$. Then for each $k,Y_k$ is a proper subspace of $Y_{k+1}$.

By Riesz’s lemma, for each $k$ there exists $x_k$ in $S=\{x∈X:||x||_{\infty}=1\}$ such that $\Vert Y_k-x_k \Vert_{\infty} ∈ (1/2, 1]$. Also, if $j ≠ k, || x_k – x_j || ≥ 1/2$. Then we have a sequence $\{x_1, x_2, x_3, ..., x_j, ... \} ⊆ S = \{x ∈ X : || x || = 1\}$ a closed and bounded set which is then compact. But since $j ≠ k$ implies $|| x_k – x_j || ≥ 1/2$, no subsequence of $\{x_1, x_2, x_3, ..., x_j, ... \}$ is Cauchy and so no subsequence can converge, contradicting the fact that $S$ is compact. As $S$ is not compact $X$ is not finite dimensional.

Note: I have looked and found posts similar to this and this, and they do not answer my question, as they do not use the above theorem/use the concept of a closed unit ball.

Answer 1

You can't really show that a space is infinite-dimensional using the theorem "$S$ compact implies $\dim X < \infty$", since the theorem doesn't say anything about what happens if $S$ is not compact. You need the converse, "$\dim X < \infty$ implies $S$ is compact" for the conclusion, and you use this converse (in the form of its contrapositive) at the end of your proof.

That said, your proof is correct, although, to be frank, a bit silly. You use the existence of an infinite linearly independent set - which immediately implies the infinite-dimensionality - to construct an infinite closed discrete subset of the unit sphere, and then conclude from that that the unit sphere is not compact and hence the space not finite-dimensional.

But since infinite linearly independent sets in $X$ are so easy to see, any proof of the infinite-dimensionality that doesn't just expose an infinite linearly independent set is a bit silly.

Here's another silly proof of infinite-dimensionality of $X$ using the theorem that a normed vector space is finite-dimensional if and only if its unit sphere is compact:

Let $\lambda \colon X \to \mathbb{R}$ be defined by

$$\lambda(f) = \int_0^{1/2} f(x)\,dx - \int_{1/2}^1 f(x)\,dx.$$

Then $\lambda$ is a continuous linear functional on $X$ ($\lvert \lambda(f)\rvert \leqslant \lVert f\rVert_\infty$ is easily seen), so if the unit sphere of $X$ were compact, then $\lambda(S)$ would be a compact subset of $\mathbb{R}$. But $\lambda(S) = (-1,1)$ is a bounded open interval, not compact. [Try proving that $\lambda(S) = (-1,1)$, it's not too hard.]