Hilbert Sequence Space is Complete Metric Space

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Theorem

Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Let $\ell^2 = \left({A, d_2}\right)$ be the Hilbert sequence space on $\R$.


Then $\ell^2$ is a complete metric space.


Proof

We have that Hilbert Sequence Space is Metric Space.

It remains to be shown that it is complete.


Recall that from Real Number Line is Complete Metric Space, $\R$ is a complete metric space.

Let $x^1, x^2, x^3, \ldots$ be a Cauchy sequence $\ell^2$.

Then for each $i \in \N_{>0}$, we have that $\left\langle{ {x_i}^j}\right\rangle_{j \mathop \in \N_{>0} }$ is a Cauchy sequence in the complete metric space $\R$.

Thus $\left\langle{ {x_i}^j}\right\rangle_{j \mathop \in \N_{>0} }$ converges to a point in $\R$, say $x_i$.

Then if $x = \left\langle{x_i}\right\rangle$, the points $x - x_j$ eventually belong to $H$.

So $x = \left({x - x_j}\right) + x^j$ must also belong to $H$.

Thus $d \left({x, x^j}\right) = 0$.

Hence the result.

$\blacksquare$


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