If $|X_k| \rightarrow 0$ then there are integers such that $m_k |X_k| \rightarrow t$ for any $t$.

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.

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