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Let $L_1$ be the set of all sequences of real numbers $$x = (x_1,x_2,..., x_n, ...) $$ with the property that $\sum_{n=1}^\infty |x_n|$ is convergent. If we define $$d_1(x,y) = \sum_{n=1}^\infty |x_n-y_n|$$ for all x and y in $L_1$, prove that ($L_1, d_1)$ is a complete metric space and the space is a Banach space in a natural way.

For complete metric spaces, I know this means to show that an arbitrary Cauchy sequence in $(L_1, d_1)$ converges to a point in $(L_1, d_1)$, but I'm not sure how to write this. $\forall \epsilon > 0, \exists N \in \Bbb N, st.$ ($\forall m,n > N \Rightarrow $ CAUCHY METRIC).

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  • $\begingroup$ I dont understand... if the series are convergent you are in a complete space. I think you must prove that $\sum|x_n-y_n|$ is a metric in this space. $\endgroup$ – Masacroso May 23 '16 at 13:20
  • $\begingroup$ The convergent series in this case refer to specific points; it doesn't necessarily mean this space is complete. To prove it is complete you must demonstrate that all cauchy sequences of sequences converges. $\endgroup$ – Neil May 23 '16 at 21:11
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Warning: this proof is full of the most disgusting indexing the world has ever known. Wait, actually in retrospect this proof has less disgusting indexing than the proof for $L_\infty$.

Suppose we have a sequence of points $x_n$. For each $x_n$, we will denote the $t^{th}$ element of the sequence-vector (recall that each point is a sequence-vector) as $x_{n,t}$. What you need to prove is that if $x_n$ is Cauchy, it converges to some $x$ in the space.

First, let us examine what it means for $x_n$ to be Cauchy. $\forall r \in \mathbb{R}^+ \exists N \in \mathbb{N}$ such that $\forall k,m > N, d_1(x_m,x_k)<r$. Now we can unpack. $d_1(x_m,x_k)=\sum_{t=1}^\infty |x_{m,t}-x_{k,t}|$.

I can continue the proof from here if you need, but it seems like you're mostly having trouble with articulating the Cauchy criterion for this sequence space.

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