Originally Posted by

**Hartlw** The Weak Link in Post 1:

Given f(x) continuous on [a,b], given ε, for each x you can find a δ > 0 st │f(x) - f(p)│ < ε if │x - p│ < δ.

Superficially, the set S of δ’s has a minimum value which proves the theorem.

Weak Link:

1) ε → 0.

2) S is infinite.

1) No matter what ε is, δ > 0.

As for 2): Houston, we have a problem. If I have an infinite set of numbers all of which are > 0, the set may not have a minimum greater than 0. For example, the set of all rational numbers whose square is > 2 does not have a minimum, it has a glb. In the case of S, a glb of 0 doesn’t work.