An uniform convergence proof of piecewise function

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I having trouble with the next function which I should show if it's uniform convergent or not for $x \in \mathbb{R} $ :

$$ f_n(x) = \begin{cases} 0 &\text{if}\quad x \leq n\\ x - n &\text{if}\quad x > n\\ \end{cases} $$

I tried to solve the problem using the $m_n = \text{sup}_{x \in \mathbb{R}} | f_n(x) - f(x)|$ sequence for proving uniform convergence. We have:

$$\lim_{n \to \infty} f_n(x) = 0 \quad \text{for all} \quad x \in \mathbb{R} $$

and then it's obvious that picking $x_n = 2n$:

$$m_n = \text{sup}_{x \in \mathbb{R}} | f_n(x)| \geq | f_n(x_n)| = n \not\to 0$$

and then we can conclude that $f_n(x)$ is not uniform convergent for all $\mathbb{R}$.

The problem I find is if can I use the sequence $x_n = 2n$, as this sequence tends to infinity and I don't know if in this case infinity can be considered as part of the domain of $f_n(x)$, I mean $\infty \in \mathbb{R}$ or not?

Or in this case the thing is that tending to infinity is not a number itself and we can just consider that all $x_n \in \mathbb{R}$?

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It of course doesn't. It converges pointwise to $0$, but $$\sup_{x\in\mathbb R}|f_n(x)|=\infty.$$