# Understanding a proof

I'm having some trouble understanding the second half of this proof that proves the following theorem:

Let f : A → R, where A ⊂ R, and suppose that c ∈ R is an
accumulation point of A. Then
limx→c
f(x) = L
if and only if
limn→∞
f(xn) = L.
for every sequence (xn) in A with xn 6= c for all n ∈ N such that
limn→∞
xn = c.

The first half of the proof: First assume that the limit exists and is equal to L. Suppose that (xn) is any sequence in A with xn 6= c that converges to c, and let  > 0 be given. From Definition 6.1, there exists δ > 0 such that |f(x) − L| <  whenever 0 < |x − c| < δ, and since xn → c there exists N ∈ N such that 0 < |xn − c| < δ for all n > N. It follows that |f(xn) − L| <  whenever n > N, so f(xn) → L as n → ∞.

The first half of this proof I follow; we simply show that for a sequence, x_n, a delta exists when the sequence converges to c, which implies an epsilon must exist to show the limit of f(x_n) = L.

What I can't understand is the following (picture used as I don't know how to format is properly):

Proof part 2

The basis of it is proving the converse of the above proof when the limit does not exist. I'm specifically stuck on where they pulled the inequality 0 < |x_n - c| < 1/n from: this doesn't seem to be a rule anywhere that I have found and I haven't the faintest how they got that result.

If anyone could give some tips on what's happening here it would be incredibly appreciated.

Given any $$\delta$$, we can always find at least one $$x \in A$$ such that $$0<|x-c|<\delta$$ and $$|f(x)-L| \geq \epsilon$$.

This means in particular that for any $$\delta_n = 1/n$$, there exists $$x_{n} \in A$$ with $$0<|x_n-c|< 1/n$$ and $$|f(x_n)-L| \geq \epsilon$$. Now $$x_n \to c$$ , but $$f(x_n)$$ remains a fixed positive distance away from $$L$$, contradicting the hypothesis that $$f(x_n) \to f(c)$$.

We stipulate that $$x_n$$ satisfies $$0<|x_n-c|< 1/n$$ for convenience. We just want to construct a sequence $$x_n$$ tending to $$c$$ whose image $$f(x_n)$$ does not tend to $$f(c)$$, and picking $$\delta_n = 1/n$$ does the trick. There is nothing stopping us from constructing a more exotic sequence of $$\delta_n$$s. But there is no need to be extravagant.

Note that this direction is nontrivial, as it involves the axiom of (countable) choice.

• So we're fixing delta to be equal to 1/n? Could this proof then also work if we fix delta = 1/20n for example?
– user431606
Commented Apr 11, 2017 at 14:12
• Yes, it still works; the detail doesn't really matter. The key idea is to construct a sequence of $x_{n}$ which get arbitrarily close to $c$, but remain a fixed positive distance away from $L$, contradicting the sequential limit hypothesis. Commented Apr 11, 2017 at 14:21
• The sequential limit hypothesis being the epsilon delta definition? As epsilon inequality is a direct result from our delta inequality? Just want to be clear on what we've contradicted as I'm kind of unsure.
– user431606
Commented Apr 11, 2017 at 14:23
• You're right. Note that in your initial comment you said "fix $\delta = 1/n$"; I just want to point out that what we're really doing is we're picking out those $\delta$s that are of the form $1/n$, $n \in \mathbf{N}$. Commented Apr 11, 2017 at 14:27
• Well the sequential limit hypothesis is the sequence definition, not the epsilon-delta one. Commented Apr 11, 2017 at 14:28