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

$$(Y,d)$$ 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.

Thanks!!

(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)

• 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. 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)

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