Understanding why an inequality holds in a proof

Question

I was working on a real analysis problem and the following problem was one that I struggled with. Here are some preliminary definitions:

A sequence of of points $\{u_{k}\}$ in $\mathbb{R}^{n}$ is Cauchy provided for each $\epsilon > 0$, there is an index $K$ such that $\text{dist}(u_{k}, u_{l}) < \epsilon$ if $k, l \geq K$.

The i-th projection function $p_{i}(u)$ returns the $i$-th component of the vector $u$. This function is continuous and linear.

And here is the problem I am working on:

Prove $\{u_{k}\}$ is Cauchy iff each component sequence is Cauchy.

Here is one solution I found somewhere. I don't understand it.

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Suppose $\{u_{k}\}$ is Cauchy. Each component sequence is Cauchy since

$$\|p_{i}(u_{k}) - p_{i}(u_{l})\| = \|p_{i}(u_{k} - u_{l})\| \leq \sqrt{\sum_{i=1}^{n}|p_{i}(u_{k} - u_{l})|^{2}} = \text{dist}(u_{k}, u_{l}) < \epsilon.$$

Here, I get the first equality is due to linearity. I don't get the second inequality though. Why is that true? BTW, we use the standard dot product as the norm. I thought that this should just be an equality, not less than or equal to.

That is the only part I don't get.

Answer 1

First of all, note the $p_i$ is a map from $\mathbb{R}^n$ into $\mathbb R$. Therefore, the first equality should be$$\bigl\lvert p_i(u_k)-p_i(u_l)\bigr\vert=\bigl\lvert p_i(u_k-u_l)\bigr\vert.$$Now, the inequality comes from$$\lvert x_i\rvert=\sqrt{\lvert x_i\rvert^2}\leqslant\sqrt{\sum_{i=1}^n\lvert x_i\rvert^2}.$$

Answer 2

The $p_i(u)$ are real numbers, so your norm is just the absolute value. The second inequality just tells you that $$ |a|\le\sqrt{a^2+\textrm{something else positive}} $$ BTW it is not a good idea to call $i$ the index in the sum, since you have used as a fixed coordinate right before and this can lead to confusion.