If components of a sequence in $\mathbb{R}^n$ converge to $0$, then so does the maximum of all its components

Question

Consider a sequence in $\mathbb{R}^K$: Let the $j$-th term in the sequence be denoted by $x_j = ((x_1)_j, (x_2)_j, \ldots, (x_K)_j)$. I've been given that for all $k = 1, 2, \ldots, K$, the sequence $(x_k)_n \to 0$ as $n \to \infty$ and I'm trying to prove that

$$\big(\max_k x_k\big)_n \to 0$$

Let $\epsilon > 0$ be given. For any $k \in \{1,\ldots,K\}$, let $N_k \in \mathbb{N}$ such that $(x_k)_n < \epsilon$ for all $n > N_k$. If $N = \max_k N_k$, then for all $k \in \{1,\ldots,K\}$, $(x_k)_n < \epsilon$ for all $n > N$, which in turn implies $\big(\max_k x_k\big)_n < \epsilon$ for all $n > N$.

Seems like a simple result to prove yet I'm not entirely sure. Does this need any corrections? Thanks

Answer 1

You may also proceed as follows:

$$|\max_{1\leq k\leq K} x_k^{(j)}| \leq \max_{1\leq k\leq K} |x_k^{(j)}| \leq \sum_{k=1}^K |x_k^{(j)}| < K \varepsilon \mbox{ for } j > N_{\varepsilon} \mbox{, if you choose } N_{\varepsilon} \mbox{ as you did.}$$

Answer 2

Alternatively, one can use that $(x_1, \dots, x_n) \to 0$ with the usual metric iff $x_i \to 0$ for $i = 1, \dots, n$.

But since the usual metric (= Euclidean metric) is equivalent with the maximum metric defined by $$d(x,y) = \max_{i=1}^n |x_i - y_i|$$

we have $(x_1, \dots, x_n) \to 0$ with the maximum metric iff $(x_1, \dots, x_n) \to 0$ with the usual metric, from which the result easily follows.

So note in particular that the converse is also true.