Supremum of bounded family of functions

Question

Let $X$ be a metric space and $\mathcal{F}$ family of functions from $X$ to $\mathbb{R}$. If $\mathcal{F}$ is bounded above in $X$, then $\sup_{f\in \mathcal{F}} f(x)<\infty$.
So far, I read in some books, definition of a family of functions $\mathcal{F}$ is bounded above in $X$ if for each $x\in X$, there exists $K_x>0$ such that $f(x)\leq K_x$ for all $f\in \mathcal{F}$.
My question: Why is $\sup_{f\in \mathcal{F}} f(x)<\infty$?
Because when I have a family of functions $\mathcal{F}=\{f_n:X\to \mathbb{R}: n\in \mathbb{N}\}$ which is defined by $f_n(x)=n$ for all $x\in X$, then for all $n\in \mathbb{N}$, $f_n$ is bounded above because for each $x\in X$, there exists $K_x=n+1$ such that $f_n(x)=n\leq n+1$. But, $\sup_{f_n\in \mathcal{F}} f_n(x)=\infty$.
Thanks for any help.

Answer 1

I think there's an understandable but critical confusion happening here. Let's look carefully at the boundedness condition you would like to apply:

A family $\mathcal{F}$ is bounded above in $X$ if, for each $x \in X$, there exists $K_x > 0$ such that $f(x) \leq K_x$ for all $f \in \mathcal{F}$.

Notice, in particular, that the bounding constant $K_x$ is allowed to depend on $x$, but it is not allowed to depend on which $f \in \mathcal{F}$ you're looking at.

Here's an example to consider: define $f_n: \mathbb{R} \to \mathbb{R}$ by $$ f_n(x) = x + \frac{1}{n} $$ for $n \in \mathbb{N}$ and set $\mathcal{F} = \{f_n\}_{n \in \mathbb{N}}$. Then $K_x = x+1$ satisfies $f_n(x) \leq K_x$ for all $x \in \mathbb{R}$, so $\mathcal{F}$ satisfies the "bounded in $X$" condition you specified (with $X = \mathbb{R}$). There are two things I'd like to point out here: first, our bound $K_x$ does not depend on our choice of $n$; second, each function $f_n \in \mathcal{F}$ is unbounded over its whole domain. Nonetheless, we happily satisfy our boundedness condition because at any point $x$ we might pick, we can easily find a number bounding $f_n(x)$ for all $n$.

The takeaway is that (perhaps counterintuitively) the "$\mathcal{F}$ is bounded above in $X$" condition is less concerned with the boundedness of any particular function over the entirety of $X$ than it is with how the entire family of functions behave, compared to one another, at a single point. Hope that helps!