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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.

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Take a rational sequence that converges to $x$. Add $\pi/n$ to each term. –  Rahul Jan 7 '13 at 6:13
    
ah, yes thanks. –  Jmaff Jan 7 '13 at 6:14
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Related question: 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 7 '13 at 6:19
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1 Answer 1

up vote 4 down vote accepted

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 $x$.

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