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 $\lim x_n = c$

  • $\begingroup$ By $R$ do you mean the real numbers $\Bbb R$? Or is $R$ some more general space? $\endgroup$ Commented Mar 5, 2019 at 22:14
  • $\begingroup$ R is indeed the real numbers. I'll make this edit. $\endgroup$
    – JaP
    Commented Mar 5, 2019 at 22:15
  • $\begingroup$ The proof is ok, except you have to choose an index $N$ such that the two inequalities you wrote are satisfied and then add and subtract $f(x_N)$. You're using that the limit of the two sequences is $c$ both because you're saying that the limits of the sequences $f(x_{n_i}})$ exist and in applying the Lemma you cited afterwards to claim that the limit of the function exists. $\endgroup$
    – Federico
    Commented Mar 5, 2019 at 23:07

1 Answer 1


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.

  • $\begingroup$ Thanks! I'll do this now. $\endgroup$
    – JaP
    Commented Mar 5, 2019 at 22:23
  • $\begingroup$ Do you mind taking a look now? $\endgroup$
    – JaP
    Commented Mar 5, 2019 at 22:46
  • $\begingroup$ If two sequences $\{f(x_n) \}$ and $\{f(y_n) \}$ go to two different limits $L_1$ and $L_2$, consider what happens to the sequence obtained when you "interleave" the two sequences: $\{f(x_1), f(y_1), f(x_2), f(y_2), \ldots \}.$ It has to be Cauchy (because since $x_n$ and $y_n$ both converge to $c$, the interleaved sequence $\{x_1, y_1, x_2, y_2, \ldots \}$ also has to converge to $c$), but the difference between $L_1$ and $L_2$ ensures that it can't be. $\endgroup$ Commented Mar 5, 2019 at 23:10
  • $\begingroup$ I got it! Thanks so much! $\endgroup$
    – JaP
    Commented Mar 5, 2019 at 23:20
  • $\begingroup$ wait, I'm still confused. Why does $ \{f(x_1), f(y_1), f(x_2), f(y_2), ... \}$ have to be Cauchy? What does that have to do with the interleaved sequence $\{x_1, y_1, x_2, y_2 \}$ being convergent? $\endgroup$
    – JaP
    Commented Mar 5, 2019 at 23:34

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