Define for a field $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}, \mathbb{F}^\omega := \{(x_n)_{n\geq 1} : x_i \in \mathbb{F}\,\forall i\}.$ Let $x=(x_n)_{n\geq 1}, y=(y_n)_{n\geq 1} \in \mathbb{R}^\omega$. Let the $1$-norm $d_1$ be defined as $d_1(x,y) := ||x-y||_1 = \sum_{i=1}^\infty |x_i-y_i|,$ where and the $\infty$-norm or sup norm to be $d_\infty(x,y) := ||x-y||_\infty = \sup_{i\in\mathbb{N}}|x_i-y_i|.$ Define $\ell_1(\mathbb{R}) := \{x=(x_n)\in \mathbb{R}^\omega: ||x||_1 < \infty\}$ and $\ell_\infty(\mathbb{C}) := \{x=(x_n)\in \mathbb{R}^\omega : ||x||_\infty < \infty\}.$ For each of the following metric spaces $A_i$, determine whether it is separable, complete, compact, and/or connected.
- $A_1 = \{a \in \ell_1(\mathbb{R}) : \|a\|_1 = 2\|a\|_2\}$ in $(\ell_1(\mathbb{R}), d_1)$.
- $A_2 = \{a \in \ell_\infty(\mathbb{C}) : |a_k| = 1\,\forall k \in \mathbb{Z}^+\}$ in the space $(\ell_\infty(\mathbb{C}), d_\infty).$
- Let $A_3 = \{a_0+a_1x+\cdots + a_nx^n : n\geq 0, a_i \in \{0,\pm 1\}\,\forall i\}$ in the space $(\mathcal{C}([0,1],\mathbb{R}), d_2),$ where $d_2(f,g) := (\int_0^1 (f-g)^2)^{1/2}$ and $\mathcal{C}([0,1],\mathbb{R})$ is the set of continuous functions from $[0,1]$ to $\mathbb{R}.$ Determine whether $A$ is separable, connected, complete, and/or compact.
I think $A_1$ is closed (it's the inverse image of $0$ under the uniformly continuous function $f(x) = \|x\|_1 – 2\|x\|_2$) and since $(\ell_1(\mathbb{R}), d_1)$ is complete, this shows $1$ is complete. It's also a subspace of the space $\ell_1$ which is separable and hence $A_1$ is separable. Since $\ell_1$ is also complete and $A_1$ is closed in $\ell_1,A_1$ is complete. I know how to show that $A_1$ is not bounded. Also, it seems the function $\alpha : [0,1] \to A_1, \alpha(x) = tx$ is a path on $A_1$.
For $A_2,$ I think it is not separable; one could define for each subset $A_2\subseteq 2^\mathbb{N}, e_{A} = (e_{A,k})_{k\geq 1}$ by $e_{A,k} = 1$ if $k\in A$ and $i$ otherwise and then obtain an injection from $2^\mathbb{N}$ to any dense subset. I also think it's closed, and since $\ell_\infty$ is complete, this'll show it's complete. But I'm not sure whether it's compact or connected.
I think the set in $A_3$ is actually countable; it's the countable union of sets $\{a_0+a_1x^1+\cdots + a_nx^n : a_i \in \{0,\pm 1\}\}$. Thus a countable dense subset could be itself. I think the sequence $f_n(x) = \sum_{i=0}^n x^i$ does not converge in the set, but I'm not sure whether this is useful as the metric space $(\mathcal{C}[0,1],d_2)$ is not complete. Also, I'm not sure whether this is compact or connected. I know that compact sets in metric spaces are totally bounded and complete, so this might be useful.
Edit: I'll add what I've come up with for the "exercise" in the last part of the answer by zhw. So we have, $|f_n-f| = \frac{x^{2n+2}}{1+x^2}\leq x^{2n+2}$ for $x\in [0,1]$ and hence $(f_n-f)^2 \leq x^{4n+4}$ so $\int_0^1 (f_n-f)^2 \leq \frac{1}{4n+5}\to 0.$
Best Answer
I think your answer for the set $A_1$ is correct.
$A_2:$ You're right that $A_2$ is not separable. This proves $A_2$ is not compact, since any compact metric space is separable.
Completeness: Suppose $a_1,a_2,\dots $ is a sequence in $A_2$ and $a_n\to a$ in $l^\infty(\mathbb C).$ Then for each $k,$ $a_n(k)\to a(k).$ This implies $|a_n(k)|\to |a(k)|$ for all $k.$ Therefore $|a(k)|=1$ for all $k,$ which says $a\in A_2.$ This implies $A_2$ is closed in $l^\infty.$ Since $l^\infty$ is complete, so is $A_2.$
Connectedness: I tend to think $A_2$ is connected, but I don't know how to prove it yet.
$A_3:$ You're right, $A_3$ is countable, hence it is separable.
$A_3$ is unbounded, hence is not compact. Proof: Let $f_n(x)= 1+x+\cdots + x^n.$ Then all $f_n\in A_3,$ and we have
$$\int_0^1f_n(x)^2\,dx > \int_0^1(1+x^2+x^4+\cdots+x^{2n})\,dx$$ $$=1+1/3+1/5+\cdots +1/(2n+1)\to \infty.$$
$A_3$ is not connected: Proof: Every connected metric space with more than one point is uncountable.
$A_3$ is not complete: Recall that for $x\in [0,1),$
$$f(x) =\frac{1}{1+x^2} =\sum_{k=0}^{\infty}(-1)^k x^{2k}.$$
For $n=1,2,\dots$ define $f_n(x)= \sum_{k=0}^{n}(-1)^k x^{2k}.$ Then each $f_n\in A_3.$ Now
$$f(x)-f_n(x) = x^{2n+2}((-1)^{n+1} +(-1)^{n+2}x^2 + (-1)^{n+3}x^4+\cdots ).$$
From this it follows that $f_n\to f$ in the $d_2$ metric. I'll leave this as an exercise for now. Since $f\notin A_3,$ we're done.