A video I was watching on the proof that a function continuous over a compact set is necessarily uniformly continuous over that set used the very intuitive idea that if $\left|x-y\right|<\delta_0:\left|f(x)-f(y)\right|<\epsilon$ then $\left|x-y\right|<\delta_1: \left|f(x)-f(y)\right|<\epsilon$, if $\delta_1<\delta_0$; this makes sense to me, since a choice of smaller $\delta$ with closer input would surely produce closer output, but I found a counter example where closer inputs have more disparate outputs:
e.g. If $f(x)=x^2$ and we choose $\epsilon$ to be $0.1$, then $x=-1, y=1$ has $\delta=2$ and a difference $\left|(-1)^2-1^2\right|=0<\epsilon$, but if I shrink $\delta$ to be $1$, then $x=-1, y=0$ and the difference in their image is now clearly $>\epsilon$.
In the context of the proof, the mathematician was assuming that if we cover the compact set with balls formed by some $x_i$ and radius $\delta(x_i)$ where these $\delta(x_i)$ satisfy a delta-epsilon continuity condition, then the smallest $\delta(x_i)$ from the set of all these deltas must also satisfy the continuity condition for all $x_i$, because a smaller delta must work for all - but that doesn't fit with my counter-example, right?
More specifically, they were saying that if (excuse any notational or formal errors - I am extremely new to these concepts) $S$ is a compact subset of the reals and $f:S\to\mathbb{R}$ is continuous over $S$:
$$\left|f(x_i)-f(x)\right|<\epsilon:\left|x_i-x\right|<\delta(x_i)$$ $$\forall x_i\in S, N\in\mathbb{N}:S\subseteq\bigcup_{i=1}^NB_{\delta(x_i)}(x_i)$$
Then:
$$\exists\delta=\min\{\delta(x_i):\forall x_i\in S\}$$
Such that:
$$\left|x_i-x\right|<\delta, \left|f(x_i)-f(x)\right|<\epsilon:\forall x, x_i \in S$$
This makes intuitive sense but I cannot reconcile it with my counter example - can anyone enlighten me?
Thanks from a complete novice to real analysis!
The idea is that if you have a $\delta_0 > 0$, such that $|f(x) - f(y)|$ for every pair of numbers $x, y$ such that $|x - y|<\delta_0$, then the same will hold if your replace $\delta_0$ by $\delta_1$ with $0 < \delta_1 < \delta_0$. (The proof is simple: if $|x - y| < \delta_1$, then $|x - y| < \delta_0$ too and we can apply what we are given about $\delta_0$.) In your example $|x^2 - y^2| < 0.1$ does not hold for every pair of numbers $x, y$ such that $|x - y| < 2$.