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Let S be a dense set in the reals. Prove that every real number is the limit of a sequence in S. I know I am supposed to consider S intersected with $(x-$ $1 \over n$, $x)$ for $n\in \mathbb N$, for a given $x \in \mathbb R$.

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What does it mean to you for a set to be dense? – anonymous Dec 9 '12 at 16:39
If a set S is dense in R, then it means that for all real a<b, there exists an x in S s.t. a<x<b – Mathlete Dec 9 '12 at 16:41
So for $x \in \mathbb{R}$ and each $n$ there's a point $x_n$ in $S$ satisfying $x - \frac{1}{n} < x_n < x$. Then $\lim_{n \to \infty} x_n$ = ? – anonymous Dec 9 '12 at 16:52
up vote 2 down vote accepted

Take any sequence $s_n \in S$ with $s_n \in (x-\frac{1}{n},x)$ for each n. What does $s_n$ tend to?

S being dense means that there is always some element of $S$ in any interval, so certainly there's an element of $S$ in $(x-\frac{1}{n},x)$ so we know that we can make such a sequence.

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Would s_n tend to x? – Mathlete Dec 9 '12 at 17:09
Yes, can you think why? – Tom Oldfield Dec 9 '12 at 17:10
Because any sequence in that interval would be bounded above and below by the values either side of the interval? – Mathlete Dec 9 '12 at 17:11
Yes, that's true, so what bound can we put on $|s_n - x|$? – Tom Oldfield Dec 9 '12 at 17:12
Can we choose an epsilon greater than 0, and say that that is less than epsilon? – Mathlete Dec 9 '12 at 17:14

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