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 $x \in \mathbb R$.
Let S be a dense set in the reals. Prove that every real number is the limit of a sequence in S.
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sequences-and-series
analysis
limits
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0So 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 $\lim_{n \to \infty} x_n$ = ? – 2012-12-09
1 Answers
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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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0Great, glad to help! – 2012-12-09