On the space of sequences with finitely many non-zero components with the supremum norm and its closed unit disc

Question

Let $X = \{ (x_1,x_2,\cdots): x_i \in \mathbb{R} \text{ and only finitely many } x_i\text{'s are non-zero}\}$ and $d: X \times X \to \mathbb{R}$ be a metric on $X$ defined by $$d(x,y) = \sup_{i \in \mathbb{N}}|x_i-y_i|$$ for $x=(x_1,x_2,\cdots)$ and $y=(y_1,y_2,\cdots)$.

$P$: $(X,d)$ is a complete metric space

$Q$: The set $\{x\in X: d(\bar{0},x) \leq 1\}$ is compact where $\bar{0}$ denotes the zero element of $X$.

Which of the following statements is true?

a) both $P$ and $Q$

b) $P$ only

c) $Q$ only

d) neither $P$ nor $Q$

My attempt: I think option a is correct because for in $X$ every Cauchy sequence converges and in $Q$ it is closed and bounded implies compact (Heine-Borel theorem)

Any hints/solution? Thank you

Answer 1

Here's my thought:

Consider the sequence

$$y_n=(1, \frac{1}{2} \cdots,\frac{1}{n},0,0,0,\cdots)$$

The sequence is Cauchy in the given metric, but the limit is not in the space. To see why it's Cauchy, note that assuming $n > m$

$$d(y_n,y_m)=\sup_{i\in\mathbb{N}}|(y_n)_i-(y_m)_i|=\frac{1}{m+1}$$

So, given $\epsilon >0$, you can choose $m$ large enough to show that the sequence is indeed Cauchy in the given metric. However, $y_n \to (1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n},\cdots)$ which is not in the space because it has infinitely many non-zero components.

For the second one, note that a compact metric space must be complete. So, if a metric space is not complete, it can't be compact. Consider $\{y_n\}$ as constructed above. Now $\{y_n\}$ is a sequence in $A=\{x\in X: d(0,x) \leq 1\}$ but it converges to $(1,\frac{1}{2},\frac{1}{3},\cdots,\frac{1}{n},\cdots)$ which is again not in the space as explained. Hence, $A$ is not compact.

Answer 2

One answer is already given. I am giving another approach for future.

Think $X$ as a normed linear space with sup norm. The set $\{ e_i : i \in \mathbb{N} \},$ where $e_i$ is an element with $1$ at $i$th position and zero elsewhere, is a countable basis. Now by Baire category theorem $P$ is false and $Q$ is also false as basis is not finite ( closed unit ball is compact iff vector sapce is finite dimensional).