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.

  • $\begingroup$ 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

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.

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