Convergence in a normed space

Question

I have that $x_n \rightarrow x$ in a normed vector space $\mathbb{X}$ and $y \in \mathbb{X}$.

I want to show that $$\|x-y\|_{\mathbb{X}} \leq \sup_{n\in\mathbb{N}} \|x_n - y\|_{\mathbb{X}}$$

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Attempt:

By definition of convergence, we have that $\forall \varepsilon > 0, \exists N \in \mathbb{N}$ such that $n>N \Rightarrow \|x_n - x\|_{\mathbb{X}} < \varepsilon$.

Then, for $n>N$ $$\|x - y\|_{\mathbb{X}} \leq \|x_n - x\|_{\mathbb{X}} + \|x_n - y\|_{\mathbb{X}} < \varepsilon + \|x_n - y\|_{\mathbb{X}}$$

I am not sure what to do from this point and I am not even sure why the supremum exists... Also the supremum in the question runs across all $n$, whereas I only have it for $n>N$?

Answer 1

What you did is fine. This proves that$$\|x-y\|_{\mathbb{X}}\leqslant\varepsilon+\sup_{n\geqslant N}\|x_n-y\|_{\mathbb{X}}.$$But note that $\sup_{n\geqslant N}\|x_n-y\|_{\mathbb{X}}\leqslant\sup_{n\in\mathbb N}\|x_n-y\|_{\mathbb{X}}$. Therefore, you proved that$$\|x-y\|_{\mathbb{X}}\leqslant\varepsilon+\sup_{n\in\mathbb N}\|x_n-y\|_{\mathbb{X}}.$$Since this takes place for each $\varepsilon>0$,$$\|x-y\|_{\mathbb{X}}\leqslant\sup_{n\in\mathbb N}\|x_n-y\|_{\mathbb{X}}.$$