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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.