# Alternate proof of "$\ell^\infty$ is not separable".

I am self-studying Functional Analysis from Kreyszig's Introductory Functional Analysis with Applications. In section-1.3, he proves that the sequence space $$\ell^\infty$$ with the metric $$d_\infty(x,y)=\sup_i\{|x_i-y_i|\}$$ is not separable.

While going through the author's proof and some other ones online, I realized that there are two main ways to show that $$\ell^\infty$$ is not separable: $$(1)$$ Showing that every dense subset of $$\ell^\infty$$ is uncountable, OR $$(2)$$ Showing that there is no countable subset of $$\ell^\infty$$ which is dense in $$\ell^\infty$$. However, I have thought about the following way of showing that $$\ell^\infty$$ is not separable that does not use either of these strategies:

Let $$Y\subset\ell^\infty$$ be the set consisting of all sequences with $$0$$'s or $$1$$'s. Then $$Y$$ is uncountable. We also notice that the restriction of the metric $$d_\infty$$ to $$Y$$ yields the discreet metric. But then by Result-$$1.3.8$$, $$\ell^\infty$$ cannot be separable since it is uncountable.

I am using the following Result from Kreyszig:

Result-$$1.3.8$$: A discreet metric space $$X$$ is separable iff $$X$$ is countable.

I feel like I am in the right direction but need more details to justify my claims. Could someone tell me if this way of showing $$\ell^\infty$$ is not separable is valid or what details to add? TIA.

What you are doing is what I would call the standard proof that $$\ell^\infty$$ is not separable. There's no real need to use any theorems or anything. It is trivial to check that $$\|y_1-y_2\|_\infty=1$$ for all $$y_1,y_2\in Y$$. As soon as you know that $$Y$$ is uncountable, you can show that any dense subset of $$\ell^\infty$$ is uncountable: because if $$D$$ is dense in $$\ell^\infty$$, for each $$y\in Y$$ there exists $$d_y\in D$$ with $$\|y-d_y\|_\infty<\frac14$$. As for each $$z,y\in Y$$ we have \begin{align} \|d_z-d_y\|_\infty&=\|(d_z-z)+(z-y)+(y-d_y)\|_\infty\\[0.3cm] &\geq\|z-y\|_\infty-\|(d_z-z)+(y-d_y)\|_\infty\\[0.3cm] &\geq\|z-y\|_\infty-\|d_z-z\|_\infty-\|d_y-y\|_\infty\\[0.3cm] &\geq1-\frac14-\frac14=\frac12, \end{align} this shows that $$d_z\ne d_y$$ if $$y\ne z$$. Thus the function $$\gamma:Y\to D$$ given by $$\gamma(y)=d_y$$ is injective, and thus $$D$$ is uncountable.

Other than misspelling "discrete" as "discreet" and writing the ambiguous "it is uncountable" rather than "$$Y$$ is uncountable", your proof is essentially correct. You might note that in a separable metric space, any family of disjoint open sets must be countable.