I'm struggling a little with this question:
Let c be a cluster point of $A ⊂ \mathbb{R}$, and $f : A → \mathbb{R}$ be a function. Suppose for every sequence
$\{x_n \}$ in A, such that $\lim x_n = c$, the sequence $\{ f (x_n )\}_{n=1}^{\infty} $ is Cauchy. Prove that $\lim_{x \rightarrow c} f(x)$ exists.
I also have the following lemma, which should probably help a lot:
Let $S ⊂ \mathbb{R}$ and $c$ be a cluster point of $S$. Let $f:S→ \mathbb{R} $ be a function.
Then $f(x)→L$ as $x→c$ if and only if for every sequence ${x_n}$ of numbers such that $x_n ∈ S – \{c \}$
for all n, and such that $\lim x_n = c$, we have that the sequence ${ f (x_n)}$ converges to L.
I feel like I'm so close but I'm missing something incredibly obvious.
Following the hint from Robert Shore, this is my attempt:
We try to prove that all sequences ${ f (x_n)} $ converge to a unique limit $L$. Assume for sake of contradiction that there are two sequences $\{x_{n_1}\} , \{x_{n_2}\}$ where $\lim \{x_{n_1}\} = \lim \{x_{n_2}\} = c$ but $\lim \{f(x_{n_1})\} = L_1 \neq \lim \{f(x_{n_2})\} = L_2$. Then from the epsilon-delta definition of limits, we have:
$$ \forall \epsilon > 0, \exists M_1, \forall n \geq M_1 \implies |f(x_{n_1}) – L_1| < \epsilon/2 $$
$$ \forall \epsilon > 0, \exists M_2, \forall n \geq M_2 \implies |f(x_{n_2}) – L_2| < \epsilon/2 $$
Then
$$ |L_1 – L_2| = |f(x_{n_1}) – L_1 – (f(x_{n_2}) – L_2 )|$$
$$ \leq |f(x_{n_1}) – L_1| + |f(x_{n_2}) – L_2|$$
$$ < \epsilon/2 + \epsilon/2 =\epsilon$$
Since $|L_1 – L_2| < \epsilon$ for all $\epsilon > 0, |L_1-L_2| = 0$ and so $L_1 = L_2$
I somehow feel like this proof is going in the wrong direction – I'm not at all using the fact that $\lim x_n = c$
Best Answer
Here's an outline of the proof. Since $\Bbb R$ is complete, every Cauchy sequence converges. Therefore, for every sequence $\{x_n \}$ converging to $c$, we have that $\{f(x_n) \}$ converges. All you need to prove is that all such sequences $\{f(x_n) \}$ converge to the same limit. (Let us know if you need help with that.) Once you know that, call that unique limit $L$ and then apply the Lemma.