Upper bound problem in real analysis

[complexnumber]

Homework Statement

Let $$\mathcal{F} \subset C(\mathbb{R})$$ be a set of continuous
functions such that for each $$x \in \mathbb{R}$$ there is an $$M_x ><br /> 0$$ such that $$|f(x)| \leq M_x$$ for all $$f \in \mathcal{F}$$.

Homework Equations

Prove that there is a nonempty open subset $$Y \subseteq X$$ and an $$M<br /> > 0$$ such that
$$\begin{align*}<br /> |f(x)| \leq M \text{ for all } x \in Y \text{ and all } f \in<br /> \mathcal{F} \text{.}<br /> \end{align*}$$

The Attempt at a Solution

I can't see the point of such a proof as isn't for any $$Y \subseteq X$$ such an $$M = \sup_{x \in Y} M_x$$? What exactly needs to be proven?

 

[Office_Shredder]

The supremum could be infinity