A similar question is asked a lot on here, but this is about why my proof is wrong.
Given two uniformly continuous functions on $\mathbb{R}$, call them $f,g$. I need to show that their composition is uniformly continuous. But the following proof -- obviously in error -- claims that the functions need not be uniformly continuous, just continuous, in order for their composition to be uniformly continuous...
Since $f,g$ are continuous on $\mathbb{R}$ their composition is continuous on $\mathbb{R}$. That is, $\forall \epsilon > 0$ we can find a $\delta(\epsilon)>0 $ such that both of the following inequalities are satisfied for some arbitrary point $c\in \mathbb{R}$
$$\left|f\left(g\left(x\right)\right) - f \left(g\left(c\right)\right)\right| < \frac{\epsilon}{2}\\ \left|f\left(g\left(y\right)\right) - f \left(g\left(c\right)\right)\right| < \frac{\epsilon}{2}$$
whenever both $x,y$ are in a $\delta $-neigborhood of $c$.
Now, why can't I, when asked to prove uniform continuity, just say okay, then for any given $\epsilon$, we will restrict the above $\delta$ to $\frac{1}{2} \delta$ so that when $x,y$ are in the $\frac{\delta}{2}$-neighborhood of $c$. The above inequalities are satisfied and by the triangle inequality we get (and letting $F = f \circ g$).
$$\left|F\left(x\right) - F\left(y\right) \right| \leq \left|F\left(x\right) - F\left(c\right)\right| + \left|F\left(y\right) - F\left(c\right)\right| < \epsilon $$
The arbitrariness of $c \implies x,y$ are arbitrary, thus giving uniform continuity for any composition of continuous functions on $\mathbb{R}$
Did I not just prove that for any given $\epsilon$ that there exists a $\delta$... my god after all that typing, I get it. My proof still depends on $c$ even though it is eliminated by the triangle quality. I could have a different $\delta$-neighborhood for each $c$. Is that right? It seems so obvious now, that it's weird.