Answer to Question #153573 in Functional Analysis for Ashweta Padhan

Question #153573

Show that Linfinity is not separable space


1
Expert's answer
2021-01-05T18:18:40-0500

The space "L^{\\infin}" is the set of all bounded sequences of real (or complex) numbers where the metric is given by "d(x,y)=sup_{n\\isin N}|x_n-y_n|" where "x=\\{x_n\\}_{n\\isin N}" and similarly for "y" .

Consider the set of all sequences that whose entries are made up of zeroes and ones. Obviously this is a subset of "L^{\\infin}" . Furthermore, each of these sequences corresponds to the binary representation of a number in "(0,1)" , and every number in "(0,1)" has a binary representation, so a bijective mapping between "(0,1)" and our set exists. This means that our set is uncountable. Note that because of the metric on "L^{\\infin}", any two (distinct) elements in the set are distance one apart. If we place a ball of radius "r<1\/2" around each point, then none of these balls will intersect. This tells us that, since any dense subset of "L^{\\infin}" must have an element in each ball, any dense subset of "L^{\\infin}" must be uncountable, so "L^{\\infin}" is not separable.


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