If the sequence $(x_n)\to a$ in $\mathbb{R}^m$ and $\|x_n\|\leq c,\;\forall n\in \mathbb{N}\;$, $\;$ then $\|a\|\leq c$

Question

Let $(x_n)_{n\in \mathbb{N}}\subset\mathbb{R}^m$ is a sequence that converges to $a\in \mathbb{R}^m$, i.e. $\lim_{n\to \infty}(x_n)=a.$

If $\|x_n\|\leq c,\; \forall n\in \mathbb{N}.$ $\;\;$ Then I want to prove that $\|a\|\leq c.$

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The only information that I have for this demonstration is: $\forall \varepsilon >0 ,\; \exists n_0\in\mathbb{N}\;\; | \;\;$ if $\;\,n\geq n_0$ then $\|x_n-a\|<\varepsilon$

and I think that something would come of of the relationship:

$\|a\|=\|x_n-a+a\|\leq \|x_n-a\|+\|a\|<\varepsilon +\|a\|$

but I get nothing, because I see how to use the hypothesis $\|x_n\|\leq c,\; \forall n$

another possible property with which I work is the reverse triangular domestic inequality:

$|\, \|x_n\|-\|a\|\, |\leq \|x_n-a\|$ then

$ \|a\|-\|x_n\|\leq \|x_n-a\|\quad \Rightarrow \quad \|a\|\leq\|x_n\| + \|x_n-a\| < \varepsilon + c $

and I fail to prove that it meets $\|a\|\leq c$.

Help please. Regards.

Answer 1

You have basically shown it! Notice that you proved that:

$$\quad \|a\| < \varepsilon + c \; \forall \; \varepsilon > 0$$

Now let $\varepsilon \searrow 0$ to get $\|a\|\leq c$.

Or if this still confuses you, also notice that the above limit is equivalent to taking

$$\inf_{\varepsilon \in (0,\infty)}\{\cdot\}$$ on both sides.

Answer 2

Hint

If $x_n\leq y_n$ then $\lim_{n\to\infty }x_n\leq \lim_{n\to\infty }y_n$.