Let $(X,d)$ be a metric space and $x_n$ be a sequence such that it has a convergent sub-sequence $x_{n_k} \to x $, then $x_n \to x$

This is false, for example, the sequence $(-1)^n$ has a convergent sub-sequence but it does not converge.

Here's a proof of the above statement, I can't figure out what the flaw is.


Suppose that $x_n$ does not converge to $x$, so there exists an $\epsilon$ such that for all $N$ we have $$n \geq N \rightarrow d(x_n, x) \geq \epsilon$$ on the other hand there is an $N_\epsilon$ such that $$n_k\geq N_\epsilon \rightarrow d(x_{n_k},x) < \epsilon$$ since $x_{n_k}$ is a subsequence of $x_n$, its members are also members of the main sequence, therefore there are members of $x_n$ such that $d(x_n,x) < \epsilon$ which is a contradiction.

What is wrong with this proof?

  • 1
    $\begingroup$ The problem is in the very first line. $d(x_n,x) < \epsilon$ for all $n\ge N$ being false does not in any way mean $d(x_n,x) \ge \epsilon$ for all $n\ge N$. It just means there is at least one $n\ge N$ where $d(x_n,x)\ge \epsilon$. $\endgroup$ – fleablood Oct 24 '19 at 22:36
  • $\begingroup$ This equivalent to the following prove. If some horse are brown than all horses are brown. Suppose all horses are not brown; then there is some other color that all horses are. Call it "grue". But some horse are brown and therefore those are not grue. That's a contradiction. So all horse are brown. $\endgroup$ – fleablood Oct 24 '19 at 22:42
  • $\begingroup$ I like that analogy, so I couldn't figure out what the flaw was because I failed to notice that the negation of convergence was wrong! How silly $\endgroup$ – clementine Oct 24 '19 at 22:49
  • $\begingroup$ When, as in this question, you have a fake proof and a specific counterexample, an effective way to find the flaw in the proof is to apply it to your counterexample. In this case, that would mean writing the proof with every $x_n$ replaced with $(-1)^n$. Then, if necessary, check line by line to see where the proof for this specific example goes wrong. $\endgroup$ – Andreas Blass Oct 24 '19 at 22:52

The contrapositive is wrong. To say $x_n \not\rightarrow x$ is to say

\begin{equation} \exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \exists n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon \end{equation}

And not

\begin{equation} \exists\epsilon>0 \text{ s.t. } \forall N \in \mathbb{N} \: \: \forall n \geq N \text{ s.t. } d(x_n,x) \geq \epsilon \end{equation}

The first allows the existence of the subsequence because not all $n$ are obligated to satisfy $d(x_{n_k},x) \geq \epsilon $.


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