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

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 $$X$$?

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?

• The fact that $\langle 2^{-n}:n\in\Bbb Z^+\rangle\notin X$ means that it is irrelevant to the question of whether $V$ is closed in $X$. – Brian M. Scott Jan 17 at 23:35
• Your solution looks fine. – Kavi Rama Murthy Jan 17 at 23:35
• @BrianM.Scott I always get confused by these examples where not only does a sequence leave the subset, but it also leaves the space entirely. Is your comment one about the fact that $X$ itself is not even closed? Every element in the sequence does lie in $V$ though. – Algebra geek Jan 17 at 23:38
• You’re working in $X$, and by definition $X$ is a closed subset of itself. Anything not in $X$ is completely irrelevant. – Brian M. Scott Jan 17 at 23:39
• Ah I see the problem. Thanks. – Algebra geek Jan 17 at 23:40

(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 $$X$$).

• Good catch! I saw the reference to (2) and just assumed it was invalid. Fixed the answer. – Pedro Amaral Jan 17 at 23:41
• You could also mention that (4) can be used to prove (3). – Brian M. Scott Jan 17 at 23:45

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 $$X$$.

• Elegant approach. – Algebra geek Jan 17 at 23:54
• @Algebrageek: Thanks. (And a good choice of which answer to accept: Pedro’s covers more ground.) – Brian M. Scott Jan 17 at 23:55
• I was doubting for a bit, but you phrased it nicely: "to complement". – Algebra geek Jan 17 at 23:56