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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.

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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.

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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.

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