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If Lp is the set of sequences of real or complex numbers, such that the infinite series, (where you sum up the moduli of the terms in the sequence to the power of p) converges to a finite value. Let the metric d(x,y) be this sum for the sequence x-y, taken to the power of 1/p. How do you prove this metric space is seperable?

user157872
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In the real case, consider the set $A=\{\sum_{j=1}^N r_j e_j | N\in\mathbb N, r_j\in \mathbb Q\}$, where $e_n$ denotes the sequence $e_n(j)=1$ if $j=n$ and zero otherwise.

If you wish, you can write $$ A=\bigcup_{n\in \mathbb N}\{\sum_{j=1}^N r_j e_j | r_j\in \mathbb Q\}. $$ Observe that we have (in the set-theoretic sense) $$ \{\sum_{j=1}^N r_j e_j | r_j\in \mathbb Q\}\simeq \mathbb Q^N, $$ and that the right-hand-side is countable. In fact, with obvious identifications, this becomes an equality, which might help you see why $A$ is dense in $\ell ^p$ for $0<p<\infty$.

In the complex case, replace $\mathbb Q$ by $(\mathbb Q +i\mathbb Q)$.

Jonas Dahlbæk
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  • But then if I just look at L1 for concreteness and consider the sequence which is zero for all terms except at the N+1th term, where it is 1. Then there is no point in that set which gets under a distance 1 from that sequence and so that sequence is not a limit point for that set. What's wrong with my reasoning? – user157872 Sep 10 '14 at 16:49
  • The element you describe is what I denote by $e_{n+1}$ (where I change notation to a lower case $n$ in order to ease notation later). This is an element of the set, corresponding to the choice $N=n+1$, $r_j=0$, $1\leq j \leq n$, $r_{n+1}=1$. – Jonas Dahlbæk Sep 11 '14 at 06:44
  • But then there is no bound on N and that set is going to become the set of all infinite sequences of rational numbers. Which is uncountable. – user157872 Sep 11 '14 at 07:32
  • Lets say we're dealing with L1, tell me what N is for that case? – user157872 Sep 11 '14 at 07:34
  • @user157872 No, the set is a countable union of countable sets, making it countable by a standard result. $N$ ranges over all positive integers. – Jonas Dahlbæk Sep 15 '14 at 20:57