Intuition behind the Epsilon-Delta definition of continuity?

Question

I understand the epsilon-delta definition of a limit but I do not understand how that relates to continuity

Definition:

Let $f:X \rightarrow \Bbb R$ be a real function, and suppose that $a \in X$ then f is continuous at a if, given any $\epsilon$ in $\Bbb R^+$, there exists $\delta \in \Bbb R^+$ such that $$\lvert f(x)-f(a)\rvert \lt \epsilon\ \text{whenever}\ x \in X \ \text{and}\ \lvert x-a\rvert \lt \epsilon$$

Would you be able to explain the intuition behind this definition? How does this show that a function is continuous?

Thanks!

Answer 1

The idea is that if you get very close to the point $a$, the values of $f$ will be very close to the values of $f(a)$.

So, if you prescribe a distance from $f(a)$ (the $\varepsilon$) then you can always find a distance (the $\delta)$ such that if $x$ is at distance less than $\delta$ from $a$, then $f(x)$ is at distance less than $\varepsilon$ from $f(a)$.

Answer 2

An intuitive approch is as a garantuee. If FOR ALL $\epsilon >0$ THERE IS SOME $\delta >0$ such that $\mid x-y \mid < \delta $ implies (or "its garantueed") that $\mid f(x)-f(y) \mid < \epsilon$

Note the importance of the "for all" and "for some", these are very important in logical statments

Answer 3

I note you may be confusing the definition of a limit and the definition of continuity.

Given a function $f:[a,b]→ℝ$ and $x∈[a,b]$, we say:

1. The limit of $f(y)$ as $y$ approaches $x$ exists and equals $L$, $\lim_{y→ x} f(y) = L$ if for any $ε>0$ there is a $δ>0$ such that if $0<|y-x|<δ$ then $|f(y) - L|<δ$.
2. $f$ is continuous at $x$ if for any $ε>0$ there is a $δ>0$ such that if $0<|y-x|<δ$ then $|f(y) - f(x)|<δ$.

In particular a function can have limits at $x$ without being defined at $x$. A function is continuous at $x$ iff $\lim_{y→ x}f(y)$ exists and equals $f(x)$.

Answer 4

I think, if I remember correctly, that this definition was an attempt by 19th century Germans to make precise the notion of a graph that can be drawn without lifting one's pencil.

Turns out that this definition doesn't quite capture that intuition, but it's a good first attempt. I'll leave it to you to ponder why this definition is a good attempt.

Answer 5

Well....

Continuous intuitively means as you go from $x$ to $w$ the values of $f(x)$ to $f(w)$ can't have a huge discontinuous jump. That means if you look at all the points really close to $f(w)$, call them $\alpha's$, they all correspond to points, call them x'es, that each x is really close to w and $f(x) = \alpha$ for some $\alpha$.

In other words. $\alpha$ really close to $f(w)$ => there is an x such that $f(x) = \alpha$ and x really close to w.

This means by making x and w really close we can "force" f(x) and f(w) to be really close. So if we want f(x) and f(w) to be "that close" we can do that by making sure x and w are "this close". (Imagine my holding my index finger and thumb together and squinting.)

Well, we've got to put this in real math term....

f(x) and f(w) being "that close" means that we can want them to be such $|f(x) - f(w)| < \epsilon$ where $\epsilon$ is an arbitrarily small number that we want to force them to be within. What we want to show is that we can force this relation by finding another small number, call it $\delta_{\epsilon}$, so that if we force x and w to be $\delta_{\epsilon}$ close to each other than it has to follow that $f(x)$ and $f(w)$ are within $\epsilon$ of each other.

(If we can not force the function to do this, than the function is "jumping" a big distance from f(x) to f(w) while x and w are a really small distance apart. This means the function isn't continuous.)

So we define a function as continuous at w if, for any small $\epsilon > 0$ we can find a $\delta > 0$ such that whenever $|x - w| < \delta$ it has to follow that $|f(x) - f(w)| < \epsilon.$