Problem: Prove the metric of $\ell_2$ satisfies the triangle-inequality
Reminders:
A point in $ \ell_2 $ space is an infinite sequence $ \mathbf x = \langle x_k \rangle_{k=1}^{\infty} $ of real numbers, for which $ \sum_{k=1}^{\infty} x_k^2 < \infty $, meaning the infinite series $ \sum_{k=1}^{\infty} x_k^2 $ converges.
The distance between $ \mathbf x = \langle x_k \rangle $ to $ \mathbf y = \langle y_k \rangle $ in $ \ell_2 $ is defined as such:
$ d(\mathbf x,\mathbf y) := \sqrt{ \sum_{k=1}^{\infty} ( x_k – y_k )^2 } $
Proof from the book: Let $ \mathbf x = \langle x_k \rangle _k , y = \langle y_k \rangle _k , z = \langle z_k \rangle _k $ be arbitrary points in $ \ell_2 $. Let $ n $ be natural number. We'll use the triangle-inequality satisfied by the Euclidian metrics and the fact that all the elements of the sums are positive so that we'll have:
$ \sqrt{ \sum_{k=1}^{n} ( x_k – z_k )^2 } \leq \sqrt{ \sum_{k=1}^{n} ( x_k – y_k )^2 } + \sqrt{ \sum_{k=1}^{n} ( y_k – z_k )^2 } \leq \sqrt{ \sum_{k=1}^{\infty} ( x_k – y_k )^2 } + \sqrt{ \sum_{k=1}^{\infty} ( y_k – z_k )^2 } = d( \mathbf x, \mathbf y ) + d( \mathbf y, \mathbf z ) $
When $ n \rightarrow \infty $ on the left-hand side of the inequality, the inequality is preserved and we get
$ d( \mathbf x, \mathbf z ) = d( \mathbf x, \mathbf y ) + d( \mathbf y, \mathbf z ) $.
My question:
In the proof from the book, I understood everything until where they've decided to take the limit $ n \rightarrow \infty $ on the left-hand side of the inequality.
I denoted $ A_n = \sqrt{ \sum_{k=1}^{n} ( x_k – z_k )^2 } $ , $ B_n = \sqrt{ \sum_{k=1}^{n} ( x_k – y_k )^2 } $ , $ C_n = \sqrt{ \sum_{k=1}^{n} ( y_k – z_k )^2 } $, I know $ A_n \leq B_n + C_n $ and also that $ A_n \leq \lim_{ k \rightarrow \infty }( B_k + C_k) $ ( since our sums are of positive numbers only ) .
However, I don't understand why the inequality when taking limit on left side : $ \lim_{ n \rightarrow \infty } A_n \leq \lim_{ n \rightarrow \infty }( B_n + C_n) $ is justified? ( because maybe $ \lim_{ n \rightarrow \infty } A_n $ does not converge or is $ +\infty $ , while $ \lim_{ n \rightarrow \infty }( B_n + C_n) $ converges to constant number ).
Note: I'm familiar with taking limits on inequalities in Real-Analysis. If we have sequences $a_{n}, b_{n}$ with $a_{n} < b_{n}$ and both the sequences converge to $a, b$ respectively then from this information we can only conclude that $a \leq b$. But the example above undermined my understanding.
Best Answer
Since $\ell_2$ is a vector space, $\sum_{k=1}^\infty(x_k-z_k)^2<\infty$. Besides, the sequence $\left(\sum_{k=1}^n(x_k-z_k)^2\right)_{n\in\Bbb N}$ is an increasing one. So (and since it is bounded) it is convergent.