Prove that $\lim_{y \to x-}f(y) = a$ if for every sequence $(x_n)_n$ in $\mathbb{R}$ that increases towards x, the sequence $f(x_n)$ converges to $a$

Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function.

Prove that: $$\forall (x_n)_n \subseteq \mathbb{R}: (x_n \to x \land \forall n: x_n \leq x_{n+1} \implies f(x_n) \to a)$$ implies that $\lim_{y \to x-}f(y) = a$

My attempt:

Suppose $\lim_{y\to x-}f(y) \neq a$. This means,

$$\exists \epsilon >0: \forall \delta >0: \exists y \in \mathbb{R}: 0 < x-y < \delta \land |f(y) - a | \geq \epsilon$$

Choosing this $\epsilon > 0$ and letting $\delta := 1/n$, we can find a sequence $(y_n)_n$ in the reals such that $0 < x-y_n< 1/n$ and $|f(y_n)-a| \geq \epsilon$

This implies that $y_n \to x$ and $f(y_n) \not\to a$ and it is also clear that $(y_n)_n$ is increasing. The result follows by contraposition.

Is this correct?

Answer 1

This is not quite right: it is not "clear" that $(y_n)$ is increasing, and in fact that may not be true. There is absolutely no reason you couldn't have $y_{n+1}<y_n$, if the only restriction in choosing $y_n$ is that $0 < x-y_n< 1/n$ and $|f(y_n)-a| \geq \epsilon$.

To fix this, you need to change how you choose the sequence $y_n$ in order to guarantee that it is increasing.

Answer 2

Attempt to fix it , correct me if wrong:

You have:

1)$0<x-y_n \lt 1/n$ and

2)$|f(y_n)-a| \ge \epsilon$.

1) Implies $\lim_{ n \rightarrow \infty} y_n =a.$

You need to choose an increasing subsequence such that

$y_{n_k} \le y_{n_{k+1}} $.

Inductively:

0) Start: $x-y_0\lt 1/n_0.$

1) Choose $n_1$ such that :

$1/n_1 \lt (x-y_0).$

Step:

2) Choose $n_{k+1}$ such that :

$1/n_{k+1} \lt (x-y_{n_k})$.

Now apply your argument for the increasing subsequence $y_{n_{k}}.$