Answer to Question #153573 in Functional Analysis for Ashweta Padhan

Question

Show that L^infinity is not separable space

Answer 1

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.