I'm going through Bert Mendelson's Introduction to Topology on my own. In fact, I've tried to go through it several times and I always get stuck somewhere. This time it's on limits. I think I understand the concept of the delta-epsilon proof. I want to show that that there is a certain range and distance such that whenever a comes within a certain range of b, f(a) comes within a certain distance of f(b). I think I finally understand that. But then I get to the exercises and I am just totally at a loss. I think I spent an hour on the first one today. If you could walk me through the thought process for this kind of proof that would be awesome, because I have no idea how to even start. Here's the first problem.
Let $X$ be the set of continuous functions $f:[a,b]\to R$. Let $d$* be the distance function on $X$ defined by $$d^*(f,g)=\int_a^b|f(t)-g(t)|dt,$$ for $f,g\in X.$ For each $f\in X$, set $$I(f) = \int_a^bf(t)dt.$$ Prove that the function $I:(X,d^*)\to (R,d)$ is continuous.
In this case, d is just the absolute value of the difference between the two numbers.
Fix $f\in X$, given $\varepsilon >0$ we need to find $\delta >0$ such that:
(1) $g\in X$ and $d^*(f,g)<\delta$ imply $|I(f)-I(g)|<\varepsilon$
Observe that, if $g\in X$:
$\left |I(f)-I(g)\right |=\left | \int_{a}^{b} f(t) -g(t) dt \right | ≤\int_{a}^{b} \left |f(t) -g(t) \right | dt=d^*(f,g)$
Hence if we take $\delta = \varepsilon$ we obtain:
$\left |I(f)-I(g)\right |≤d^*(f,g)<\delta=\varepsilon$
This means that $I$ is continuous at $f$, since condition (1) is met.
Since $f \in X$ was arbirtrary, $I$ is continuous in $X$.