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Let $D$ be a set of càdlàg functions on $[0,1]$.

Define $f_\alpha(\cdot) \equiv 1(\cdot \ge \alpha)$ for $\alpha \in [0,1]$, which is obviously in $D$.

Then, if we denote $|| \cdot ||$ as a uniform metric, then $||f_\alpha - f_\beta||=1$ for any $\alpha \neq \beta$.

I learned this implies non-separability of $D$ space, but I failed to complete the proof.

Who can give me any hint or something? Thanks!

Davide Giraudo
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inmybrain
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1 Answers1

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If $(D,\|.\|)$ were separable, any subset would be separable, in particular $\{f_{\alpha}\}_{\alpha\in [0,1]}$. But it's discrete and non-numerable, so it's not separable.

Nahuel Albarracín
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    Oh, I didn't know that a subset of separable set is also separable. Thanks! – inmybrain Dec 14 '14 at 12:24
  • One who is curious of that, refer to http://math.stackexchange.com/questions/516886/prove-that-a-subset-of-a-separable-set-is-itself-separable – inmybrain Dec 14 '14 at 12:26