# metric spaces with certain topological properties

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.

1. $$A_1 = \{a \in \ell_1(\mathbb{R}) : \|a\|_1 = 2\|a\|_2\}$$ in $$(\ell_1(\mathbb{R}), d_1)$$.
2. $$A_2 = \{a \in \ell_\infty(\mathbb{C}) : |a_k| = 1\,\forall k \in \mathbb{Z}^+\}$$ in the space $$(\ell_\infty(\mathbb{C}), d_\infty).$$
3. 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 $$\int_0^1 (f_n-f)^2 \leq \frac{1}{4n+5}\to 0.$$

• Please define $l^1(\mathbb R)$ and the other spaces. Also note that "1 is closed" doesn't make much sense.
– zhw.
Jun 30, 2021 at 3:01
• @zhw I clarified the definitions. Jun 30, 2021 at 13:03
• Are you looking for progress on the exercise as stated in the quote (i.e., check separability, connectedness, completeness, compactness for $A_1, A_2, A_3$), or just for verification of the arguments you've given yourself? Jun 30, 2021 at 13:06
• @silver what quote are you talking about? Also, many of my "arguments" are very incomplete, so I'd say I'm looking for solutions for the problems. Jun 30, 2021 at 16:16
• Gotcha. With "quote" I meant the thing you formatted as a quote, i.e., "Define for a field [...] , and/or connected." Jun 30, 2021 at 16:57

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.

• thanks for your efforts. If you figure something new out, could you update your answer? Also, I don't think $|a_m(k)-a_n(k)| = 2$ necessarily in your second case, but I'm pretty sure it's closed and this can be shown using the fact that $|a(k)| \leq |a(k)-a_n(k)| + |a_n(k)|$ and $|a(k)-a_n(k)| \leq ||a-a_n||_\infty \to 0$. Jul 1, 2021 at 1:32
• Also, for proving that $A_3$ is disconnected, I thought of trying to show that the map $P\mapsto P(0)$ is continuous and that the image of $A_3$ under it is clearly disconnected, but I'm not sure how to show that map is continuous. Though your proof should still work. I tried proving $A$ is not closed, but I'm not sure how to do this. Also I think $A_2$ is connected but I'm not sure how to show it. Jul 1, 2021 at 1:40
• You're right, I messed up on the completeness of $A_2.$ It's much easier than what I had. Now edited, thank you.
– zhw.
Jul 1, 2021 at 2:28
• I added the proof that $A_3$ is not complete.
– zhw.
Jul 1, 2021 at 15:02
• can you verify my solution to your "exercise" at the last part of your answer? Jul 1, 2021 at 15:24