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Let $X_n$ be a cauchy sequence not constant in a complete metric space $M$

Prove/Disprove: The limit point of $A=\{X_n\} \in M$.

Let us assume that the limit point of $A=\{x_n\} \not \in M$. Therefore $X_n\rightarrow x$ but $x\not \in M$ so $X_n$ is a cauchy sequence which does not converges which mean that the metric space $M$ is not complete, contradiction

Is the proof valid? am I missing some points or formality?

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    $\begingroup$ I think you should just show it directly. Let $\{X_n\}$ be a cauchy sequence in $M$. Since $M$ is complete, $X_n\to x\in M$. Thus $x$ is the only limit point of $\{X_n\}$ and $x\in M$. $\endgroup$ – TSF May 6 '18 at 14:29
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There's a small problem with your proof: it assumes that there are limit points. But since $M$ is complete, that's trivial. Indeed, there's only one limit point.

But my main objection is: it makes no sense to assume that the limit of the sequence doesn't belong to $M$. After all, $M$ is the whole space.

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  • $\begingroup$ Edited, the question was written with the mistake $\endgroup$ – gbox May 6 '18 at 14:24
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    $\begingroup$ @gbox I've add another sentence to my answer and another word to your question. $\endgroup$ – José Carlos Santos May 6 '18 at 14:30

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