0
$\begingroup$

I've come across a statement in lecture notes which uses the following statement:

If $|X_k| \rightarrow 0$ then there are integers $m_k$ such that $m_k |X_k| \rightarrow t$ for any $t \in \mathbb{R}$.

This doesn't make sense to me as I would have thought that the expression: $m_k |X_k| \rightarrow 0 $ as $k \rightarrow \infty$
Starting like:

$$\forall \epsilon > 0, \exists N \in Z: n > N \implies |X_k| < \epsilon $$ Now then want to show that given t, there are a sequence of $m_j$ such that $$|m_n|X_n|-t|<\epsilon$$ I can't seem to get anywhere.

$\endgroup$
  • $\begingroup$ The statement is false in general. Consider $X_k = 0$ for all $k$. $\endgroup$ – Ayman Hourieh Oct 24 '19 at 12:13
  • $\begingroup$ Are you sure there is no hypothesis on $X_k$ ? $\endgroup$ – nicomezi Oct 24 '19 at 12:13
3
$\begingroup$

The only extra assumption you need is $X_k \neq 0$ for $k$ sufficiently large.

Let $\epsilon >0$ and consider the interval $(\frac {t-\epsilon} {|X_k|},\frac {t+\epsilon} {|X_k|})$. The length of this interval exceeds $1$ for $k$ sufficiently large and hence it contains an integer $m_k$. We have $|m_k|X_k| -t| <\epsilon$ for $k$ sufficiently large.

EDIT: there is a problem with this proof because the choice of the integers $m_k$ depends on $\epsilon$. But it is easy to avoid this problem. Just replace $\epsilon$ by $\sqrt {|X_k|}$ in the argument.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.