How to prove that, if a sequence of integers converges, then its limit is an integer?

Suppose sequence is $z_n$ and limit is $z$. This is same as proving if $\forall\epsilon>0, $ there exists a $N$ s.t. $n>N$ implies $|z_n-z|<\epsilon$

I have no idea how to start. I was trying to use contradiction and suppose that limit is not an integer, and then pick $\epsilon $ that makes the limit an integer. But I don't know how to make inequality $|z_n-z|<\epsilon$ into equality.

  • $\begingroup$ If $z$ is not an integer, you can find $\epsilon > 0$ such that $(z-\epsilon, z+\epsilon)$ does not contain any integer. $\endgroup$ – Rigel May 7 '17 at 16:13
  • $\begingroup$ For example, if the limit is $2.37$, then infinitely many terms of the sequence lie on the interval $(2.36 , 2,38)$ $\endgroup$ – ThePortakal May 7 '17 at 16:16


You can prove this by contradiction.

Choose $\epsilon = \min(z-\lfloor z \rfloor,\lceil z\rceil - z)$.

Is it possible to find $z_n \in \mathbb{Z}$ such that $|z_n-z|< \epsilon$?

  • $\begingroup$ Why do we need $\frac{\min(z-\lfloor z \rfloor,\lceil z\rceil - z)}{2}$ instead of just $\frac{(z-\lfloor z \rfloor)}{2}$? $\endgroup$ – 1412 May 7 '17 at 16:59
  • $\begingroup$ For a number like $0.9$, If we only consider the floor function, $\epsilon = 0.45$, $|1-0.9| < 0.45$. Remark, actually we do not need to divide by $2$. I shall remove the denominator. $\endgroup$ – Siong Thye Goh May 7 '17 at 17:05

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