Is my proof in calculus correct?

Question

My Math professor asked me to prove a theorem in Calculus.
After some thinking, I came up with a simple proof, but I’m not sure if it's correct.

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The Challenge

Given $X^i = (X^i_1,\dots,X^i_n)$, prove the following theorem: $$\text{If } X^k \to X \text{ then } X^k_j \to\ X_j \text{ for every }\ j = 1,2,\dots,n. $$

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My proof

1. If $ X^k \to X$ then $\forall \epsilon > 0\exists n \in \mathbb{N}$ so that $\forall k \geq n$, $\|X^k-X\| < \epsilon$.

2. Let us assume there's some $j$ for which $X^k_j \to\ X_j$ is false.

3. It follows that $X^k - X = (0,0,\dots,A,0,\dots,0)$ where $A = |X^k_j - X_j|$.

4. $|X^k - X| = \sqrt{0^2+0^2+\dots+A^2+0^2+\dots+0^2} = \sqrt{A^2} = A$.

5. $ A > \epsilon$.

6. From stages 4 and 5, it follows that $\|X^k-X\| > \epsilon$. Contradiction!

7. Therefore, $X^k_j \to X_j$ Q.E.D.

My proof

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Thanks in advance!

Answer 1

Your proof is correct.

A straight forward proof is

If $ Xk→X$ then, $ ∀ϵ>0∃n∈N$ so that $∀k≥n, ∥Xk−X∥<ϵ.$

Note that for every j,we have $$|Xkj−Xj|\le ∥Xk−X∥<ϵ.$$.

Answer 2

This might come in handy. If $X$ is a vector, you have $$|X_k| \le \|X\|, \qquad 1 \le k \le n.$$