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
f(x) = L
if and only if
f(xn) = L.
for every sequence (xn) in A with xn 6= c for all n ∈ N such that
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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – user431606
    Commented Apr 11, 2017 at 14:12
  • $\begingroup$ 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. $\endgroup$ Commented Apr 11, 2017 at 14:21
  • $\begingroup$ 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. $\endgroup$
    – user431606
    Commented Apr 11, 2017 at 14:23
  • $\begingroup$ 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 $1/n$, $n \in \mathbf{N}$. $\endgroup$ Commented Apr 11, 2017 at 14:27
  • $\begingroup$ Well the sequential limit hypothesis is the sequence definition, not the epsilon-delta one. $\endgroup$ Commented Apr 11, 2017 at 14:28

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