Why do we show $<\epsilon$

Question

When we prove things like continuity in real analysis, why do we always aim for the result $<\epsilon$ when any positive multiple of $\epsilon$ proves the same result?

Answer 1

Well, first of all, that's not always what we do, because as you say, $<c\epsilon$ is enough, where $c\in(0,\infty)$.

However when you're taking a real analysis class (especially for the first time!) we want to make sure that what's obvious to the professor is also obvious (or at least understandable) to the student. So we sometimes insist on $<\epsilon$ for that reason.

P.S. Some people do it for what they believe is simplicity or ease of reading.

Answer 2

We could use an alternative definition, consider the following alternative definition of convergence when it sufficies to prove for any positive multiple K of epsilon:

We say $x_n \rightarrow l$ when $(\exists K > 0)(\forall \epsilon > 0)(\exists N)(\forall n > N) |x_n - l| < K\epsilon$.

Then consider the proposition that if $x_n \rightarrow l, y_n \rightarrow l_2$ then $x_n + y_n \rightarrow l + l_2$.

Then we have for some $K$ and $K_2$, consider any $\epsilon > 0$ $$(\exists N)(\forall n > N) |x_n - l| < K\epsilon$$ $$(\exists N_2)(\forall n > N_2) |y_n - l_2| < K_2\epsilon $$

Then consider $M = N + N_2$:

$$(\exists M)(\forall n > M) |x_n + y_n - (l + l_2)| < (K+K_2)\epsilon $$

This means that the positive multiple is $K + K_2$ so the proposition holds.

The issue is that this is really ugly compared to just choosing half epsilon.