Proof $B(X,Y)$ is Banach if $Y$ is Banach I know this has been asked a million times but I'm having trouble with a proof provided by my professor. He defined $B(X,Y)$ being the space of bounded functions (not necessarily a linear transformation). We start with $(f_n) \subset B(X,Y)$ a Cauchy sequence. Given $x \in X$, we have that
$$
\| f_n(x) - f_m(x) \| \leq \| f_n - f_m\|_{\infty} < \epsilon,
$$
where $\| f_n \|_{\infty} = \sup_{x \in X} \| f_n(x)\|$. From this, we conclude that $(f_n(x))$ is a Cauchy sequence in $Y$, so it has a limit point $f(x)$ that defines a function $f\colon X \to Y$. We want to show that $f \in B(X,Y)$. Fixed $n_0 \in \mathbb{N}$, we have that
$$
\| f(x) \| = \| f(x) - f_{n_0}(x) + f_{n_0}(x) \| \leq \| f(x) - f_{n_0}(x) \| + \|f_{n_0}(x)\| \\
\leq  \|f - f_{n_0}\| + \|f_{n_0}\|
$$
and this concludes that $f \in B(X,Y)$. My question is how do I know that $\| f - f_n \|$ doesn't go to infinity? Also, to show that $f_n \rightrightarrows f$, he states that
$$
\| f - f_n \|_{\infty} = \lim_{m \to \infty} \| f_m - f_n \|_{\infty}
$$
Why is that?
Thanks in advance!
 A: $\|f(x)\|=\lim \|f_n(x)\| \leq \sup_n \|f_n\|$. $(\|f_n\|)$ Is a Cauchy sequence of real numbers so it is bounded. ($|\|f_n\|-\|f_m\|| \leq \|f_n-f_m\| \to 0$ as $n, m \to \infty$).
Similarly, $\|f(x)-f_n(x)\| \leq \sup_{n \geq m} \|f_m(x)-f_n(x)\|$ for any $m$.  Now use Cauchy property to show that $\|f-f_n\| \to 0$.
A: The last step in
$$
\| f(x) \| = \| f(x) - f_{n_0}(x) + f_{n_0}(x) \| \leq \| f(x) - f_{n_0}(x) \| + \|f_{n_0}(x)\| \\
\leq  \|f - f_{n_0}\| + \|f_{n_0}\|
$$
goes too far. Instead, you can note that $(f_n)$ is Cauchy, and from
$$
\big|\,\|f_n\|-\|f_m\|\,\big|\leq\|f_n-f_m\|
$$
you can conclude that the number sequence $\{\|f_n\|\}$ is Cauchy, so bounded. So there exists $c>0$ with $\|f_n\|\leq c$ for all $n$. Then
$$
\|f(x)\|\leq\|f(x)-f_n(x)\|+\|f_n(x)\|\leq \|f(x)-f_n(x)\|+c\|x\|. 
$$
Taking limit as $n\to\infty$ you get $\|f(x)\|\leq c\|x\|$ and so $f\in B(X,Y)$.
And now, given $\varepsilon>0$, if $m$ is big enough so that $\|f_n-f_m\|<\varepsilon$ for all $n\geq m$,
\begin{align}
\|f(x)-f_m(x)\|&\leq\|f(x)-f_n(x)\|+\|f_n(x)-f_m(x)\|\\[0.3cm]
&\leq\|f(x)-f_n(x)\|+\|f_n-f_m\|\,\|x\|\\[0.3cm]
&\leq\|f(x)-f_n(x)\|+\varepsilon\,\|x\|. 
\end{align}
As this works for all $n$ big enough, taking limit  you get
$$
\|f(x)-f_m(x)\|\leq\varepsilon\,\|x\|
$$
for all big enough $m$, which is
$$
\|f-f_m\|<\varepsilon.
$$
So $\|f-f_m\|\to0$ as $m\to\infty$.
