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I have a few questions about the following proof taken from https://math.berkeley.edu/~sarason/Class_Webpages/solutions_202B_assign11.pdf.

Prove that the closed unit ball of an infinite-dimensional Banach space is not compact.

Proof. Let $B$ be an infinite-dimensional Banach space. Choose any unit vector $x_1\in B$, and let $A_1$ be the subspace spanned by $x_1$. In the quotient space $B/A_1$, choose a coset of norm $\frac{1}{2}$ and then a representative $x_2$ of that coset of norm at most 1. Then $\|x_2−x_1\| \geq \frac{1}{2}$. Let $A_2$ be the subspace spanned by $x_1$ and $x_2$,and note that $A_2$ is closed. In the quotient space $B/A_2$, choose a coset of norm $\frac{1}{2}$ and then a representative $x_3$ of that coset of norm at most 1. Then $\|x_3−x_2\| \geq \frac{1}{2}$ and $\|x_3−x_1 \| \geq \frac{1}{2}$. Continuing in this way, we obtain a sequence $\{x_n\}$ of vectors in the closed unit ball of $B$ such that $\|x_m−x_n\| \geq \frac{1}{2}$ whenever $m\neq n$. The sequence $\{x_n\}$ then has no convergent subsequences, implying that the closed unit ball in $B$ is not compact.

My questions are:

1) How do we know that a coset with a norm $\frac{1}{2}$ exists in $B/A_1$?

2) Why does the closedness of $A_2$ matter?

3) Why does it matter if $B$ is a Banach space?

Teodorism
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3 Answers3

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You may want to review the definition and properties of quotients of a normed space.

  1. Because $B / A_1$ is a normed space, and it is not the zero space unless $A_1 = B$, which is not the case here ($B$ is infinite-dimensional and $A_1$ is one-dimensional). So it contains at least one nonzero element (remember elements of $B / A_1$ are cosets), and you can multiply this by an appropriate scalar to obtain something of norm $1/2$.

  2. If you take a quotient of a normed space by a non-closed subspace, in general you only get a seminormed space (the resulting "norm" may fail to be positive definite). Since we want $B / A_2$ to actually be a normed space (see step 1), we need $A_2$ to be closed.

  3. It doesn't matter. It is also true that "the closed unit ball of an infinite-dimensional normed space is not compact".

Nate Eldredge
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    Re 3., it does matter in the sense that for an infinite-dimensional normed-but-not-Banach space we have a much shorter proof than for Banach spaces. If the space is not complete, its closed unit ball isn't complete either, thus not compact. – Daniel Fischer Oct 06 '19 at 10:51
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Your method is similar to the result of the famous Riesz Lemma but not the same. (See https://en.wikipedia.org/wiki/Riesz%27s_lemma) An important result proved by Riesz Lemma is that the unit ball of an infinite-dimensional normed space X is never compact.

Yuanchuan Li
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You can similarly consider $l^2$, the (Hilbert) space of square-summable sequences $\{a_n\}$ with $\Sigma_{i=1}^{\infty} a_i^2 < \infty $ and $|| a_n -b_n||_2:=\Sigma_{i=1}^{\infty} (a_i -b_i)^2$. Then consider the close and bounded set $\{e_i\}:= \delta_i^j$, i.e., equal to $1$ in the ith spot and 0 elsewhere. This is the unit ball in $l^2$. Then , for any $i \neq j , ||a_i-a_j ||=\sqrt 2 $, so it is impossible for any subsequence to converge, as it cant even be Cauchy.

MSIS
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