# Is the set of all binary sequences compact in $l^{\infty}$?

Here, the metric space is defined as the set of all bounded sequences, with the distance function defined as the supremum of the absolute value of the difference between corresponding elements.

Intuitively, it seems that this set is not compact, although my intuition for compactness isn't very strong. I've been trying to show that the set doesn't contain all its limit points and is therefore not closed, and hence not compact, but I'm not sure how to proceed. I'm able to show that no element of this set is a limit point, but I'm not sure how to handle limit points outside of the set.

• @AndrésE.Caicedo The set of binary sequences is also bounded, so how does that observation help? Oct 5 '18 at 23:55
• @pokerlegend23 One common approach is to look for a sequence (of binary sequences, in this case) that has no convergent subsequence. Oct 5 '18 at 23:56
• @AndrésE.Caicedo yes, but the norm on $\ell^\infty$ is typically given by $$\|(x_n)\| = \sup_{n \in \Bbb N} |x_n|$$ so that even your arbitrarily long stretch of $1$s has distance $1$ from $\vec 0$. Oct 5 '18 at 23:59
• @pokerlegend The set of binary sequences is, however, a closed subset of $\ell^\infty$. So, trying to show that the subset is non-compact because it doesn't contain its limit points will inevitably fail Oct 6 '18 at 0:02
• @Omnomnomnom Oops. :-) Oh, well. Oct 6 '18 at 0:28

Consider the collection of open sets $$\mathcal{B}:=\left\{B_{l^\infty}(x,1): x \in \{0,1\}^{\mathbb N} \right\}$$. Notice that for $$x, y \in \{0,1\}^{\mathbb N}$$ we have $$\left\| x - y \right\|<1$$ if and only if $$x=y$$. So $$\mathcal B$$ covers $$\{0,1\}^{\mathbb N}$$ but there is no way to refine $$\mathcal B$$ to a finite subcover.

Note: $$\{0,1\}^{\mathbb N}$$ denotes the set of all binary sequences.

• Can an open cover be an uncountable collection of sets? Oct 6 '18 at 0:21
• @pokerlegend23 I'm not the OP but yes: an open cover can be indexed by an arbitrary set. However, if your space is a metric space, it will be compact if and only if countable covers have finite subcovers. That is, you can restrict to the countable ones to prove compactness. To prove that a space is not compact, though, it may be of use to have an arbitrarily large cover. Oct 6 '18 at 0:24
• Yes it can. Thanks for the informative response Guido A :) Oct 6 '18 at 0:30
• Note that the binary sequences form a closed and discrete subset of the whole space. So very non-compact: the only discrete compact space is a finite one. Oct 6 '18 at 4:52
• @HennoBrandsma you read my mind. If we remove just one of the open balls from $\mathcal B$, then the resulting collection of sets no longer forms a cover of $\{0,1\}^{\mathbb N}$. Oct 6 '18 at 5:26

Let $$(x^{(n)})_{n \geq 1} \subset l^\infty$$ defined by

$$x^{(n)}_i = \chi_{[n,+\infty)}(i) = \cases{0 \text{ if } i < n \\ 1 \text{ otherwise}}$$

Now, if the space of binary sequences were compact, this sequence should have a convergent subsequence $$(x^{(n_k)})_{k \geq 1} \to x$$. In particular, we should have that

$$\chi_{[n_k,+ \infty)}(i) = x^{(n_k)}_i \xrightarrow{k \to +\infty} x_i$$

and so necessarily, $$x \equiv 0$$. However, the original subsequence does not converge to zero, since

$$d(x^{(n_k)},0) = \sup_{i \geq 1}x^{(n_k)}_i = 1 \quad (\forall n \in \mathbb{N}).$$

Let $$e_n$$ be the binary sequence with $$1$$ in the $$n$$th slot and zeros everywhere else. If $$m\ne n,$$ then $$\|e_m-e_n\|_{l^\infty} = 1.$$ Thus $$(e_n)$$ has no convergent subsequences. It follows that the set of binary sequences is not compact in $$l^\infty.$$