$(Y,d)$ is a complete metric space $\rightarrow$ $(B(X,Y),p)$ is a complete metric space.

Where $B(X,Y)$ is the space of all bounded functions from $X$ to $Y$ and $p$ is the sup norm given by:

$p(f_n,f_m) = sup\{d(f_n(x),f_m(x)) : x \in X \}$

Here is the first part of the proof, up until the part that I am struggling with:

Suppose $f_n(x)$ is a cauchy sequence in $B(X,Y)$. Then, given any $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that $p(f_n,f_m) < \epsilon$ whenever $n,m > N$. This implies that

$sup\{d(f_n(x),f_m(x)) : x \in X \}<\epsilon$

So that :

$d(f_n(x),f_m(x)) < \epsilon$ whenever $n,m > N$, $\forall x \in X$

As such, $f_n(x)$ is a cauchy sequence in $Y$ for each $x \in X$.

Since $(Y,d)$ is a complete metric space, $f_n(x)$ converges to a unique point, $f(x)$, in $Y$, for each $x \in X$.

Observe that $f$ is a function from $X$ to $Y$. (This is the line I would like clarification on).

I mean, $f_n(x)$ is simply a sequence of points in $Y$, albeit a sequence defined by a sequence of functions. How do we know that this sequence of points in $Y$ converges to a point that is defined by some function $f: X \rightarrow Y$? Also, how do we know it is the same function for every $x \in X$? I mean, it seems suprising to me, considering that for each $x \in X$, $f_n(x)$ will be a different sequence.


(For anyone interested, after this step, all one has to do to finish the proof is show that $f$ is bounded, which isn't very hard)

  • 5
    $\begingroup$ it should be $f_n$ is a Cauchy sequence in $(B(X,Y),p)$, not $f_n(x)$. You define $f\colon X\to Y$ by $f(x)=\lim f_n(x)$ for all $x$. In any reasonable set theory this is a function. $\endgroup$ May 26 '19 at 1:32

As mentioned in the comments, your starting statement "Suppose $f_n(x)$ is a Cauchy sequence in $B(X,Y)$" should have been "suppose $\{f_n\}_{n=1}^{\infty}$ is a Cauchy sequence in $B(X,Y)$". Aside from that, what you have done is correct; you used completeness of $Y$ to show that for every $x \in X$, the sequence of points $\{f_n(x)\}_{n=1}^{\infty}$ is a Cauchy sequence in $Y$; hence $\lim \limits_{n\to \infty} f_n(x)$ exists, is an element of $Y$ and is unique (since limits in metric spaces are unique).

So, to each $x \in X$, we can assign a unique element $f(x) \in Y$ by the rule $f(x) = \lim \limits_{n \to \infty}f_n(x)$.

This is exactly what it means to define a function (to each element of the domain, you assign a unique element of the target space). So indeed, we have a function $f: X \to Y$.

To finish up the proof, as you said, you have to show that the function $f$ defined above is bounded (so that it is actually an element of $B(X,Y)$) and you also have to show that $f_n \to f$ with respect to the metric $p$, i.e show that

For every $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \in \mathbb{N}$, if $n > N$, then $p(f_n, f) < \varepsilon$.

(these last two being pretty easy to prove)

  • $\begingroup$ The the function $f$ that $f_n$ converges to is defined by where it maps each $x$, cool. Thanks man. $\endgroup$
    – Num Toez
    May 26 '19 at 20:34

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