Proving cauchy criterion for limits

[k3k3]

Homework Statement

Prove the converse of the Cauchy Criterion for Limits.

Let I be an interval that either contains the point c or has c as one of its endpoints and suppose that f is a function that is defined on I except possibly at the point c. Then the function f has limit at c iff for all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}

Homework Equations

The Attempt at a Solution

Already proven is:

If for all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}, then f has limit at c.

I want to show:

If f has limit at c, then or all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}

Suppose the limit of f as x tends to c is L.

Since f has limit at c, then or all ε>0 there exists a δ>0 such that |f(x)-L|<ε for all x,y in I that satisfy |x-c|<δ

Then |f(x)-f(y)+f(y)-L|≤|f(y)-f(x)|+|f(y)+L|

Since f has limit as c, let S = {t in I| 0 < |t-c| < δ}

Choose x,y from S

Then |f(y)-f(x)|+|f(y)+L| < |f(y)-f(x)| + ε

This is where I am stuck. I want to show that |f(y)-f(x)| < ε.

 

[jbunniii]

I want to show:

If f has limit at c, then or all ε>0 there exists a δ>0 such that |f(y)-f(x)|<ε for all x,y in {t in I| 0 < |t-c| < δ}

Suppose the limit of f as x tends to c is L.

Since f has limit at c, then or all ε>0 there exists a δ>0 such that |f(x)-L|<ε for all x,y in I that satisfy |x-c|<δ

Then |f(x)-f(y)+f(y)-L|≤|f(y)-f(x)|+|f(y)+L|

This is true but not the inequality you want. Try |f(y) - f(x)| = |f(y) - L + L - f(x)| \leq |f(y) - L| + |f(x) - L| \ldots