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For every real $x$, is there an irrational sequence that converges to $x$?

I thought this was true for some reason. I know it's true that there always exist such rational sequences and there's already a post about that. Thanks.

Michael Albanese
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Jmaff
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    Take a rational sequence that converges to $x$. Add $\pi/n$ to each term. –  Jan 07 '13 at 06:13
  • ah, yes thanks. – Jmaff Jan 07 '13 at 06:14
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    Related question: http://math.stackexchange.com/q/81759/. As a modification of Rahul's comment, if $x$ is irrational, the sequence $(x,x,x,\ldots)$ will do. If $x$ is rational, then $(x+\pi,x+\pi/2,x+\pi/3,\ldots)$ will do. (Thus density of the rational numbers need not be used.) – Jonas Meyer Jan 07 '13 at 06:19

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Let $x$ be real. For each positive integer $k$, there is an integer $m_k$ such that $x-\frac{1}{k}<\frac{m_k}{k+1}<x+\frac{1}{k}$. Now the rational sequence $(\frac{m_k}{k+1})_k$ converges to $x$, so the irrational sequence $(\frac{m_k+\sqrt 2}{k+1})_k$ converges to $x$.

Alternatively, if $x$ is rational, then the rational sequence $(x,x,x,...)$ and the irrational sequence $(x+\frac{\sqrt 2}{k})_k$ converge to $x$. If $x$ is irrational, then the irrational sequence $(x,x,x,...)$ converges to $x$.