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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0What does it mean to you for a set to be dense? – 2012-12-09
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0If a set S is dense in R, then it means that for all real a – 2012-12-09
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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
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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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0Would s_n tend to x? – 2012-12-09
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0Yes, can you think why? – 2012-12-09
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0Because any sequence in that interval would be bounded above and below by the values either side of the interval? – 2012-12-09
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0Yes, that's true, so what bound can we put on $|s_n - x|$? – 2012-12-09
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0Can we choose an epsilon greater than 0, and say that that is less than epsilon? – 2012-12-09
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0@Mathlete We can if we choose $n$ large enough. We say that $s_n \rightarrow x$ if $|s_n - x| \rightarrow 0$ as $n \rightarrow \infty$. Since $s_n \in (x - \frac{1}{n}, x)$, $|s_n - x| <\frac{1}{n}$. Hence for any $\epsilon > 0$ picking $N$ such that $N > \frac{1}{\epsilon}$ we have that for all $n \geq N$, $|s_n - x| < \frac{1}{n} < \frac{1}{N} < \epsilon$. Hence $|s_n - x| \rightarrow 0$ so $s_n \rightarrow x$. Does that help? – 2012-12-09
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0I completely understand that now, thanks. – 2012-12-09
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0Great, glad to help! – 2012-12-09