A subset of the vector space of all real sequences with finitely-many non-zero elements. Boundedness, closedness, compactness and completeness.

Question

Let $X$ be the space of all real sequences with finitely many nonzero elements, equipped with the supremum norm. That is, for each $x= (x_k)_{k \in \mathbb N}$ we have: $$ \|x \| = \sup \{ |x_k| k\in \mathbb N \}.$$ Consider the subset $V$ of $X$ defined by: $$ V:= \{ x= (x_k)_{k \in \mathbb N} \in X | \sum_{k=1}^\infty |x_k| \leq 1 \}.$$

I wish to answer the following practice exam questions:

1. Is $V$ bounded in $X$?
2. Is $V$ closed in $X$?
3. Is $V$ compact in $X$?
4. Is $V$ complete in $X$?

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1. Consider $x=(x_n)_{n \in \mathbb N}\in V$, we then have that for each $n\in \mathbb N$ we may write: $$ |x_n| \leq \sum_{k=1}^\infty|x_k| \leq 1 $$ Each term is nonnegative and therefore the entire sum is certainly more than a single term. We thus find that $1$ is an upper bound for each element of the sequence hence certainly it is larger than or equal to the least upper bound: $$ \| x\| \leq 1. $$ We conclude that $V$ is bounded.
2. I think this is not true as there are sequences in $V$ that leave the space (and $X$). Consider $x^{(n)}$ given by: $$x^1= (\frac{1}{2}, 0, 0, \dots) $$ $$x^2=(\frac{1}{2}, \frac{1}{4}, 0, \dots) $$ $$x^3=(\frac{1}{2}, \frac{1}{4}, \frac{1}{8},0, \dots) $$ Which might be described as $(x^n_k)_{k \in \mathbb N}$ with $x^n_k=\frac{1}{2^{k}}$ for $k\leq n$ and $0$ otherwise ($0 \not \in \mathbb N$). Each of the elements of this sequence lives in $X$ and since the geometric series converges to $1$ also in $V$. So $$ \sum_{k=1}^\infty |x^k_n| \leq 1 .$$ However, we run into a problem since its limit is $x=(x_k)_{k\in \mathbb N}$ with $x_n=\frac{1}{2^n}$, which does not have finitely many nonzero elements hence it is not in $X$ and then certainly not in $V$.
3. Compact normed vector spaces are closed, since the space is not closed, it cannot be compact.
4. The space cannot be complete as the example in (2) is a Cauchy-sequence, but the sequence does not converge to a limit in $V$. In a complete space every Cauchy sequence is convergent (with limit in the space).

Did I do this okay?

Answer 1

(1) Your solution looks correct.

(2) To show that $V$ is not closed in $X$ you need to find a sequence of elements in $V$ that converge to an element in $X$. Your sequence contains only elements of $V$ but it does not converge to an element in $X$. For that reason it is not a valid proof.

(3) Your proof relies on (2) which is not proven. You could use (4) to prove this (Hint: Is it possible for a space that is not Cauchy complete to be compact?)

(4) This proof is correct since being Cauchy complete is a property inherent to the space and does not depend on the ambient space $X$ (this is in contrast to the property of being closed which does depend on $X$).

Answer 2

To complement Pedro Amaral’s answer, I’ll show that $V$ is closed in $X$.

Suppose that $x=\langle x_n:n\in\Bbb Z^+\rangle\in X\setminus V$, so that $\sum_{n\ge 1}|x_n|>1$. Let

$$\epsilon=\sum_{n\ge 1}|x_n|-1>0\,.$$

If $y\in X$, and $\|y-x\|<\epsilon$, then $|y_n-x_n|<\epsilon$ for each $n\in\Bbb Z^+$, and therefore $|y_n|>|x_n|-\epsilon=1$ for each $n\in\Bbb Z^+$. Therefore $y\notin V$, and the $\epsilon$-ball centred at $x$ is disjoint from $V$. This shows that $V$ is closed in $X$.