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This is homework. The problem was also stated this way:

Let A be a dense subset of $\mathbb{R}$ and let x$\in\mathbb{R}$. Prove that there exists a decreasing sequence $(a_k)$ in A that converges to x.

I know:

A dense in $\mathbb{R}$ $\Rightarrow$ every point in $\mathbb{R}$ is either in A or a limit point of A.

If x is a limit point of A, then there is a sequence in A that converges to x.

What if $x\in A$?

Also, how can I know if the sequence is increasing or decreasing?

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    That would depend on what you mean by "from any direction". If you mean "along any line", then certainly not. Removing a line from a dense set does not make it not dense.2012-09-12

4 Answers 4

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There is a sequence converging to $x$, but you won't know it's decreasing - you'll have to construct it so this happens.

Try thinking about what would happen if this weren't true - then there would be an $a_0 \in A$ so that $(x,a_0) \cap A = \emptyset$... in other words this interval is full of points that are in $\mathbb{R}$, but not in $A$. Would such an $A$ still be dense?

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Hint: You have to construct your sequence so it is decreasing. Let $a_0$ be a point in $A$ greater than $x$. How do you know there is one? Then let $a_1$ be a point in $A$ in $(x,a_0)$. How do you know there is one? Then keep going. You also have to make sure the intervals shrink to zero length.

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Recall that $A \subset R$ is dense if and only if every open subset of $\mathbb{R}$ contains an element of $A$.

Let $x \in \mathbb{R}$. We will define a decreasing sequence in $A$ converging to $x$ as follows: Consider the open interval $(x + 2^{-1}, x + 2(2^{-1}))$. Since $A$ is dense, $A$ intersect this interval. Choose $a_1 \in A \cap (x + 2^{-1}, x + 2(2^{-1}))$. By recursion, suppose $a_1, ..., a_n$ has been chosen. Consider the interval $(x + 2^{-n}, x + 2(2^{-n})$. Again since $A$ is dense, there exists a $a_{n + 1} \in A \cap (x + 2^{-n}, x + 2(2^{-n})$.

The sequence $(a_k)$ constructed in this way is decreasing and converges to $x$.

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Hint: Recursively define $a_n \in A$ such that $a_n \in (x, a_{n-1})$.