Why is L∞L^{\infty} not separable?

lp(1p<) and Lp(1p<) are separable spaces.

What on earth has changed when the value of p turns from a finite number to ?

Our teacher gave us some hints that there exists an uncountable subset such that the distance of any two elements in it is no less than some δ>0.

Actually I don't understand the question very well, but I hope I have made the question clear enough.

Thank you in advance.

Answer

To be separable means to have a countable dense subset. Suppose that (M,d) is a metric space and that UM be an uncountable subset and r>0. Suppose that for all xyU, d(x,y)r. Let C be any countable subset of M. Then C can only meet a countable number of the balls Br/2(x) for xU. Let G be the union of all Br/2(x) for xU that do not meet C. G is a nonempty open subset of M that does not meet C.

There can be no countable dense subset of M.

Consider the case of . For each subset Q of the integers, let xQ be the sequence that is 1 on G and 0 off of it. The xQ are uncountable and any two elements of this collection are distance 1 apart. We have just shown that is not separable.

You can generate a similar construct for L. Consider the uncountable subclass of characteristic functions {χBr(0)}r>0L(Rn). Then each pair of distinct elements in it would be 1 unit distance apart. Ergo there cannot be any countable subset of L(Rn) that is dense in it.

Attribution
Source : Link , Question Author : Andylang , Answer Author : user3342072

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