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  • I am trying to construct a proof that continuous function do not preserve Cauchy sequences

  • Every proof I can find is disprove by counter example, which is great but these counter examples cannot be extend to how we can fix the failure of Cauchy preservation through uniform continuity.

Prove: Let $(x_n)$ be a Cauchy sequence on a metric space $(M,d)$, let $f: (M,d_M) \mapsto (N,d_N)$ be a continuous function, then $(f(x_n))$ may not be Cauchy.

(Note: I worded the problem myself which maybe a little bit awkward, feel free to point out a good fix)


(Use definition + by contradiction)

Let $(x_n)$ be a Cauchy sequence on $(M,d_N)$. Then $(x_n)$ is Cauchy if $\forall \epsilon > 0, \exists N \in \mathbb{N}$, such that $\forall n, m \geq N, d_M(x_n, x_m) < \epsilon$

Let $f$ be a continuous function such that $f: (M,d_M) \mapsto (N,d_N)$. Then given $x_o \in M, \forall \epsilon>0, \exists \delta(\epsilon,x_o) > 0 \text{ s. t. } \forall x \in M, d_M(x,x_o) < \epsilon \implies d_N(f(x), f(x_o)) < \delta $

Suppose $(f(x_n))$ is Cauchy on $(N,d_N)$, then $\forall \epsilon > 0, \exists N' \in \mathbb{N}$, such that $\forall n', m' \geq N', d_N(f(x_{n'}), f(x_{m'})) < \epsilon$

At this point I thought I would be notice some inconsistencies between the definition thus revealing a problem but I am not confident with my conclusions...

  • Can we say that we reach a contradiction because $f$ cannot be continuous since it isn't true for $\forall x$ since $(x_n)$ is only Cauchy for $n \geq N$?

  • Can we say that since $M$ is not complete, therefore $(x_n)$ may fail to convergence, thus $f(x_n)$ is not defined as $n \to \infty$?

What is the problem here exactly?

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    $\begingroup$ A conter-example is of course great to disprove the statement. There is no "generalization" of what you ask, because some cauchy sequence are preserved, and some other are not, and we can't do better. By the way, in a complete metric space, your statement is true ! $\endgroup$
    – Surb
    Mar 25, 2016 at 7:59
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    $\begingroup$ How can there be a "proof" since sometimes continuous functions preserve cauchy sequences and sime others not? Unless you are trying to find the exact point in the proof where continuity doesn' get you any further (which may be done of course). $\endgroup$ Mar 25, 2016 at 8:54

2 Answers 2

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If $d$ is a complete metric on $X$ and $f:X\to X$ is continuous then any Cauchy sequence $(p_n)_{n\in N}$ has a limit point $p$, and the continuity of $f$ requires that $(f(p_n))_{n\in N}$ converges to $f(p).$

So to find an exception we must use an incomplete metric.

Example. Let $X$ be the positive reals with the usual metric $d(x,y)=|x-y|$. Let $f(x)=1/x.$ The image of the Cauchy sequence $(1/n)_{n\in N}$ is the sequence $(n)_{n \in N}.$

It is not necessary that the metric be unbounded. In the above example, replace the metric $d$ with the equivalent metric $e(x,y)=\min (1, |x-y|).$

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  • $\begingroup$ Hi thanks for the counter example, but I can't help but think of if $X = \mathbb{R}$ instead, wouldn't the image of the sequence $(1/n)$ under $f(x) = 1/x$ also be...non-cauchy? It would still be $n$ wouldn't it $\endgroup$
    – Fraïssé
    Mar 25, 2016 at 20:51
  • $\begingroup$ But in R the function f(x)=1/x for x>0 is not continuous no matter what value you choose for f(0). The usual metric on R is complete so a continuous f on R will preserve the Cauchy property. $\endgroup$ Mar 25, 2016 at 21:02
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The statement says that $(f(x_n))$ may not be Cauchy.

The keyword here is "may not", hence trying to prove by contradiction (in the manner you described) is not possible, as the negation of "may not" is still "may not".

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