0

I am trying to negate the following statement:

Let $\{x_n\}$ in a metric space $M$ and let $p\in M$. if for all $\varepsilon > 0$ there is $k$ such that $x_k\neq p$ and $d(x_k,p)<\varepsilon$ then $\{x_n\}$ has a sub-sequence that converges to $p$

My attempt: let $\{x_n\}$ in a metric space $M$ and let $p\in M$. if $\{x_n\}$ has no a sub-sequence that converges to $p$ then there is $\varepsilon > 0$ and there is $k$ such that $x_k = p$ or $d(x_k,p)\geq\varepsilon$

Is it correct?

newhere
  • 3,115

1 Answers1

1

It is not correct. You have change de order and the quantifiers are worng.

The following statement is the correct negation:

Let $\{x_n\}$ in a metric space $M$ and let $p\in M$. If there exists an $\varepsilon>0$ such that for all $k$ with $x_k\neq p$ and $d(x_k,p)\geqslant \varepsilon$, then $\{x_n\}$ has no sub-sequence that converges to $p$.

Correction

The following statement is the correct negation:

Let $\{x_n\}$ in a metric space $M$ and let $p\in M$. If $\{x_n\}$ has no sub-sequence that converges to $p$, then there exists an $\varepsilon>0$ such that for all $k$ with $x_k\neq p$ and $d(x_k,p)\geqslant \varepsilon$.

GoRza
  • 429
  • 2
  • 15
  • 1
    ֲif the statement is of the form $p \to q$, then the negation is $\neg q \to \neg p$ ? – newhere May 19 '21 at 10:41
  • You are totally correct, sorry for that. Thanks, I've edited – GoRza May 19 '21 at 10:46
  • why negation of $x_k \neq p$ and $d(x,p)<\varepsilon$ is not $x_k = p$ or $d(x,p)\geq \varepsilon$ ? – newhere May 19 '21 at 15:45
  • 1
    Beacuse '$x_k\neq p$' doesn't depend on $\varepsilon$. Look that if so, if $x_k=p$ for all $k$, obviusly the sequence has a subsequence that converges to $p$, but it would satisfy your statement. – GoRza May 19 '21 at 20:56