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I have the following prove which I don't understand from a certain point:

Let be $ \mathcal{B}(X,\mathbb{R}) $ the set of bounded functions and $ \|\cdot\|_{\infty} $ the uniform norm. Then $ (\mathcal{B}(X,\mathbb{R}),\|\cdot\|_{\infty}) $ is a banach space.

The prove (until I got stuck):

Let be $ (f_n)_{n\in \mathbb{N}} $ a cauchy sequence in $ \mathcal{B}(X,\mathbb{R}) $. Then there exist for all $ \varepsilon>0 $ an $ N_{\varepsilon}\in \mathbb{N} $ such that for all $ m,n\geq N_{\varepsilon} $ $ \|f_n-f_m\|_{\infty}<\varepsilon $.

This implies $ |f_n(x)-f_m(x)|<\varepsilon $ for all $ x\in X $ and for all $ m,n\geq N_{\varepsilon} $. (*)

For all $ x\in X $ $ (f_n(x))_{n\in \mathbb{N}} $ is a real cauchy sequence in $ \mathbb{R} $. The fact that $ \mathbb{R} $ is complete implies that for all $ x\in X $ the cauchy sequence $ (f_n(x))_{n\in \mathbb{N}} $ converges which means $ \lim\limits_{n\to \infty} f_n(x)=:f(x) $ does exist. Now we have to show:

1.) $ (f_n)_{n\in \mathbb{N}} $ converges uniformly to $ f $ with respect to $ \|\cdot\|_{\infty} $ and

2.) $ f $ is in $ \mathcal{B}(X,\mathbb{R}) $.

So far so good. Now here comes the part I don't understand:

1.) Let be $ \varepsilon>0 $ and $ x\in X $ arbitrary. Then it is with (*)

$$ |f_n(x)-f(x)|=\left|f_n(x)-\lim\limits_{m\to \infty} f_m(x)\right|\\\ \stackrel{|.|\text{ is continous}\\}{=}\ \lim\limits_{m\to \infty} |f_n(x)-f_m(x)|\leq \varepsilon $$

Why we get in the end $ \leq \varepsilon $ instead of $ <\varepsilon $??

hallo007
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    Taking the limit can give non-strict inequalities. E.g. $1 - \frac{1}{n} < 1$ for all $n \in \Bbb Z_+$ but clearly $\lim_{n \to \infty}(1 - \frac{1}{n}) = 1 $. – balddraz Dec 01 '20 at 01:18
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    One way of looking at it is that $X=(-\infty, \epsilon)$ is open and if $x_k \in X$ and $x_k \to x$ all we can conclude is that $x \in \overline{ (-\infty, \epsilon) } = (-\infty,\epsilon]$. – copper.hat Dec 01 '20 at 01:53

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