$|a_n-\alpha|<2\epsilon$ valid as a definition of convergence?

My question is concerning the proof below taken from Ivan Wilde’s notes on real analysis which are freely available on the web.

Is coming within $\large k\epsilon$ [$k\in\mathbb Z$] of a point as effective a proof of convergence as coming within $\large \epsilon$ of it?

For k$\epsilon$ can be made arbitrarily small.
Against - (and specifically regarding the question below) - since the reasoning always comes out at 2$\epsilon$ it cannot ever meet the definition of convergence

• but then you could just substitute and redefine the variable at the last step.
• is there a meaningful analogy between this and one of the better known proofs of the irrationality of $\surd 2$ whereby the numerator of $\frac{p}{q}$ is always even, and it is that which provides the contradiction.

THEOREM
Suppose that $(a_n) [n ∈ N]$ is a sequence such that $a_n$ → α and also $a_n$ → β as n → ∞.
Then α = β, that is, a convergent sequence has a unique limit.

PROOF Let ε > 0 be given.

Since we know that $a_n$ → α, then we are assured that eventually $(a_n )$ is within ε of α. Thus, $\exists\;N_1 ∈ N: n > N_1\implies |a_n - α|<ε$

Similarly, we know that $a_n$ → β as n → ∞ and so eventually $(a_n)$ is within ε of β. Thus, $\exists\; N_2 ∈ N: n > N_2\implies |a_n - α|<ε$

So far so good. What next? To get both of these happening simultaneously, we let $N = max { N_1 , N_2 }$. Then n > N means that both $n > N_1$ and also $n > N_2$ . Hence we can say that if n > N then both $| a_n − α | < ε$

and

$| a_n − β | < ε$.

Now what? We expand out these sets of inequalities. n > N (for example n = N + 1 would do). Then

Pick and fix any

$α − ε <a_n < α + ε$

$β − ε < a_n < β + ε$

The left hand side of the first pair together with the right hand side of the second pair of inequalities gives $α − ε < a_n < β + ε$ and so

$α − ε < β + ε$.

Similarly, the left hand side of the second pair together with the right hand side of the first pair of inequalities gives β − ε < a n < α + ε and so

$β − ε < α + ε$.

Combining these we see that

$−2ε < α − β < 2ε$

which is to say that $\large| α − β | < 2ε$.

This happens for any given ε > 0 and so the non-negative number $| α − β |$ must actually be zero. But this means that α = β and the proof is complete.

Solution 1:

Yes, that method is perfectly fine. What you want to show is it works for any $\epsilon>0$. Showing it works for any $k\epsilon$ where $k>0$ is exactly the same, just by rescaling the $\epsilon$ in question. You can easily prove that is an if and only if.

If one wants to make the proof "cleaner" you can go back in the steps to find where you need to rescale things to end up with $\epsilon$ and not $2\epsilon$, but it is usually more work than it is worth

Solution 2:

This is a nice question because this sort of argument must be confusing to many beginning calculus students. Thus, it's not a bad idea to check formally that this works. In fact, let $f\colon (0,+\infty) \to (0,+\infty)$ be any surjection. Then I claim that the following is equivalent:

1. $(\forall \varepsilon > 0 )(\exists N\in\mathbb N)\ n \geq N \implies |a_n-\alpha| < \varepsilon,$
2. $(\forall \varepsilon > 0 )(\exists N\in\mathbb N)\ n \geq N \implies |a_n-\alpha| < f(\varepsilon).$

Proof. (1. implies 2.) Assume that 1. is true and let $\varepsilon > 0$. Then, $f(\varepsilon) > 0$ so if we apply 1. to $f(\varepsilon)$ instead of $\varepsilon$, there exists $N\in\mathbb N$ such that $n\geq N \implies |a_n-\alpha|<f(\varepsilon)$. Since $\varepsilon$ was arbitrary positive real, we have proven 2.

(2. implies 1.) Assume that 2. is true and let $\varepsilon > 0$. Let $\varepsilon'>0$ such that $f(\varepsilon')=\varepsilon$ (by surjectivness of $f$) so if we apply 2. to $\varepsilon'$ instead of $\varepsilon$, there exists $N\in\mathbb N$ such that $n\geq N \implies |a_n-\alpha|<f(\varepsilon') = \varepsilon$. Since $\varepsilon$ was arbitrary positive real, we have proven 1. $\square$

In their answer, Alan writes the following:

If one wants to make the proof "cleaner" you can go back in the steps to find where you need to rescale things to end up with ϵ and not 2ϵ

This is precisely looking for $\varepsilon' > 0$ such that $f(\varepsilon') = \varepsilon$ in the above proof.

Perhaps surprisingly, $f$ doesn't even have to be surjective onto $(0,+\infty)$. It is enough that the infimum of the range of $f$ is $0$. The proof is almost the same, the only change is that you would want $\varepsilon'>0$ such that $0<f(\varepsilon') < \varepsilon$, which is guaranteed to exist by $0$ being the infimum of $f(0,+\infty)$.

Solution 3:

It is totally correct actually you can fix $ε>0$;

then when applying the definition of convergence in coming steps set $ε'=ε/2>0$ then you will have at the end $ε/2+ε/2=ε$

actually you can prove easily that for a given function of epsilon $f(ε)$ such that if epsilon goes to 0 , then $f(ε)$ goes to 0 . if you have $(∃N∈N) n≥N⟹|a_n−α|<f(ε).$

then $(∀ε>0)(∃N'∈N) n≥N'⟹|a_n−α|<ε.$