Is the subset of $\ell^1$ where the $k$th coordinate is $0$ or $2^{-k}$ perfect?

Question

I've encountered the following problem:

Determine whether or not the following set is perfect in a metric space $(X,d)$: $\prod_{k=1}^\infty \left\{0, \frac{1}{2^k} \right\}$ in $(\ell_1, \|\cdot\|_1)$.

The problem is that I don't really understand this notation. First of all, how does the sequence above exactly work? I'm reading it as follows:

$$\prod_{k=1}^\infty \left\{0, \frac{1}{2^k} \right\}=\left\{0, \frac12\right\}\times \left\{0, \frac14\right\}\times\dots \times \left\{0, \frac1{2^k} \right\}\times\dots$$

But what kind of a sequence is this? How does the norm on this sequence work? Also, isn't it true by definition that sequences in $\ell_1$ have the metric $\|\cdot\|_1$?

Answer 1

As said in comments, this is the set $C$ of all sequences $x$ such that for each $k$, the entry $x_k$ is either $0$ or $2^{-k}$. The metric is inherited from the space $\ell^1$, which already has a metric induced by $\|\cdot \|_1$ norm.

The set $C$ is

• Closed. Indeed, if $y\notin C$, then there is $k$ such that $y_k$ is neither $0$ nor $2^{-k}$. Hence, any sequence $z$ sufficiently close to $y$ will also have $z_k\notin \{0,2^{-k}\}$. This shows the complement of $C$ is open.
• Totally bounded. For any $\epsilon>0$ there is $N$ such that $2^{-N}<\epsilon$, and this means $\sum_{k>N}2^{-k}<\epsilon$. So we can group the sequences in $C$ into $2^N$ subsets according to the first $N$ entries of the sequence; each subset has diameter $< \epsilon$.
• Without isolated points. If $x\in C$ and $\epsilon>0$, there is $k$ such that $2^{-k}<\epsilon$. Flip the $k$th entry of $x$ from $0$ to $2^{-k}$ or back, and we get $y\in C$ such that $\|x-y\|<\epsilon$.

In a Banach space, closed and totally bounded implies compact (since a closed subset of a complete metric space is complete).