Can $(f(x_{n}))$ always has a Cauchy subsequence?

Question

By definition, if $f:\mathbb{R}\rightarrow\mathbb{R}$ is not continuous, then there is a discontinuity point $x$. Then, $\exists \{x_n\}\rightarrow x\ s.t.\ f(x_n)\nrightarrow f(x)$.

My question is the following. From the sequence $\{f(x_n)\}$, can we find a subsequence $\{f(x_{n})\}_m$ such that $\{f(x_{n})\}_m$ is cauchy or equivalently is convergent. Can this statement always hold?

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My guess is that we can. But I don't have any clue to prove it. Can anyone help me prove it? Thanks in advance.

Answer 1

If the sequence is bounded, you can just apply Bolzano-weierstrass and use the fact that cauchy <=> convergent in $\mathbb{R^d}$

Answer 2

Let's consider classical signum function: $$f(x)=\begin{cases} -1, & x<0,\\ 0, & x=0,\\ 1 & x>0 \end{cases}$$ Then, in point $x=0$ function have discontinuity, but if we consider sequence $x_n=\frac{1}{n}$, then $x_n \to 0$ and $f(x_n)=1\to 1$ is constant i.e. converged i.e. Cauchy sequence, so, such is any its subsequence.

Answer 3

Let $(X, d)$ be a metric space. Then $A\subset X$ is called totally bounded or pre compact if $\forall\epsilon>0 ,\exists \{a_1, a_2, \ldots, a_n\}\subset A$ such that $$A\subset \bigcup_{i=1}^{n} B(a_i, \epsilon) $$

Theorem $1$: $A\subset (X, d) $ is totally bounded iff every sequence in $X$ has a Cauchy subsequence.

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Let $f:(X, d) \to (Y, d') $ be a function which is discontinuous at $a\in X$.

Then $\exists (x_n) \subset X$ such that $f(x_n) \not\to f(a) $

Claim: $(f(x_n))\subset Y$ has a Cauchy subsequence if the set $\{f(x_n):n\in \Bbb{N}\}$ is totally bounded.

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Theorem $2$ : $A\subset (\Bbb{R}^n, d_{\text{euclidean}}) $ is totally bounded iff it is bounded.

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(Ignore the previous discussion if you don't know metric space)

$f:\Bbb{R}\to \Bbb{R}$ is discontinuous at $a\in\Bbb{R}$.Then $\exists (x_n) \in \Bbb{R}$ such that $x_n\to a $ but $f(x_n) \not \to f(a) $

If the set $\{f(x_n) :n\in\Bbb{N}\}$ is bounded then $(f(x_n))$ has a cauchy subsequence.

In case of euclidean space, it's easy to prove (bounded sequence has convergent subsequence and convergent subsequence is cauchy).

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Examples:

1. $f:\Bbb{R}\to \Bbb{R}$ defined by $$f(x) =\begin{cases}1&x\ge 0\\0&x<0\end{cases}$$

Take $x_n=\frac{1}{n}\to 0$ .Then $\{f(x_n):n\in\Bbb{N}\}=\{1\}$

All subsequence of $(f(\frac{1}{n})) $ is Cauchy.

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2. $f:\Bbb{R}\to \Bbb{R}$ defined by $$f(x) =\begin{cases}\frac{1}{x}&x\neq 0 0\\0&x=0\end{cases}$$

Take $x_n=\frac{1}{n}\to 0$ .Then $\{f(x_n):n\in\Bbb{N}\}=\{n:n\in\Bbb{N}\}$.

$(f(\frac{1}{n})) $ has no Cauchy subsequence.

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