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The subspace of null sequences $c_0$ consists of all sequences whose limit is zero. Prove that $c_0$ is a closed subspace of $C$ (The space of convergent sequences), and so again a Banach space.

There's something I don't understand. I know we have to prove that every Cauchy sequence on $c_0$ is convergent on $C$ in order to prove $c_0$ is closed on $C$. But, that Cauchy sequence will be a sequence of sequences? Because the elements of $C$ and $c_0$ are sequences. I'm really confused.

Emil
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    I like to think of elements in $c$ and $c_0$ as infinitely long vectors. It's not a totally rigorous way of looking at things but it keeps you from getting your wires crossed like this. In practice, you treat them as though they were infinitely long vectors so there's not a lot of harm in it. Just know that rigorously, the individual elements are actually sequences. An alternative way to look at them is by their graph on the $xy$-plane. Their graph is just a set of discrete points where the $x$ values are restricted to the natural numbers. – Cameron Williams Jun 11 '15 at 23:53
  • I answer this here a while ago, here is the link http://math.stackexchange.com/questions/1210006/is-this-proof-that-c-0-is-a-closed-subspace-of-ell-infty-correct-also-nee/1212081#1212081 – Alonso Delfín Jun 12 '15 at 00:02
  • @AlonsoDelfín thanks! – Emil Jun 12 '15 at 14:50
  • @Emil you are welcome! I hope it is clear enough, although you want to prove $c0$ is closed in $C$ and my prove does it in $\ell^\infty$, simply note that $C \subset \ell^\infty$. Let me know if it is anything you don't understand by putting a comment over there ! – Alonso Delfín Jun 12 '15 at 14:56
  • @Emil Also in this question http://math.stackexchange.com/questions/1276470/proving-ellp-is-complete/1276508#1276508 , not the same one as yours though, but in the first part of my answer I provide an explanation on how to see sequence of sequences, which it may be your trouble also, please take a look – Alonso Delfín Jun 12 '15 at 15:11
  • Alternatively, you can think of elements of $c_0$ as functions from $\mathbb{N}$ to $\mathbb{C}$. Now the argument will look very similar to, say, the proof that $C_0(\mathbb{R})$ is complete. – Prahlad Vaidyanathan Jun 16 '15 at 07:34

2 Answers2

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Yes, you're looking at a sequence of sequences. The “distance” between two bounded sequences $(x_1,x_2,\ldots)$ and $(y_1,y_2,\ldots)$ is given by the so-called $\ell^{\infty}$-norm, which is defined as $$\|(x_1,x_2,\ldots)-(y_1,y_2,\ldots)\|_{\infty}\equiv\sup_{n\in\mathbb N}|x_n-y_n|.$$ The Cauchy property and convergence of a sequence of sequences should be evaluated with respect to this norm.

triple_sec
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$c_0$ is a vector subspace of the closed space $c$, so you just have to prove:

If $x_n \rightarrow x$ where $x_n \in c_0$ and $x\in c$, then $x\in c_0$.

mathematician
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