MT2002 Analysis

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The epsilon-delta definition

The above definition of convergence using sequences is useful because the arithmetic properties of sequences gives an easy way of proving the corresponding arithmetic properties of continuous functions. We now use this definition to deduce the more well-known epsilon-delta definition of continuity.

Informally: Close points (delta apart) are mapped to close points (epsilon apart).

We formalise this to get the following:

Definition

A function f from R to R is continuous at a point p belongs R if
given epsilon > 0 there exists delta > 0 such that if |p - x| < delta then |f(p) - f(x)| < epsilon.

Theorem
This definition is equivalent to the previous one.

Proof
First, we show that continuity in this definition implies continuity in the previous definition.
So suppose (xn) rarrow p. We must prove that (f(xn)) rarrow f(p).
That is given epsilon > 0 we must find N such that |f(xn) - f(p)| < epsilon if n > N.
Given such an epsilon from the new definition of continuity there is a delta such that |f(xn) - f(p)| < epsilon whenever |xn- p| < delta.
Since (xn) converges to p, there is a N such that this happens whenever n > N and so with this value of N our definition is satisfied.

Secondly, we show that the sequential definition implies the epsilon-delta one.
Given epsilon > 0, suppose that we could not find a suitable delta. Then delta = 1 would not work and so we must have some x1 such that |x1- p| < 1 and |f(x1) - f(p)| > epsilon.
Similarly, delta = 1/2will not work, and so we can find x2 further down the sequence than x1 such that |x2- p| < 1/2 and |f(x2) - f(p)| > epsilon.
Continuing in this way we get a sequence (x1, x2, x3, ... ) which by construction converges to p, but for which f(xn) is always at least epsilon away from f(p). So we cannot have (f(xn)) converging to f(p) and we have a contradiction.


Application

The function defined by f(x) = sqrtx is continuous.

Proof
Given epsilon > 0 we must show that |sqrtx - sqrtp| < epsilon provided that x, p are close enough.
Now |sqrtx - sqrtp| = |x - p|/|sqrtx + sqrtp| < |x - p| /sqrtp and so choosing delta = epsilon/sqrtp will do.


Remark

The definition of continuity "doesn't quite work" at p = 0 for this function since the function is not defined for x < 0. One can define a notion of "one-sided continuity" to take care of examples like this.



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JOC September 2002