Let S be contained in $\displaystyle \mathbb{R} $ and let A be the set of accumulation points of S. Prove S is dense in $\displaystyle

\mathbb{R} $ iff. A=$\displaystyle

\mathbb{R}$

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- Mar 18th 2010, 08:36 PMpirouetteAccumulation points, dense set
Let S be contained in $\displaystyle \mathbb{R} $ and let A be the set of accumulation points of S. Prove S is dense in $\displaystyle

\mathbb{R} $ iff. A=$\displaystyle

\mathbb{R}$ - Mar 19th 2010, 06:22 AMFocus
What is your definition of dense? Is it that for each ball in the reals, it contains a point of S?

If so then suppose that S is dense. Let $\displaystyle x \in \mathbb{R}$, then for each open set around x, it contains an element of S. Hence $\displaystyle x \in A$. As x was arbitrary, we have that $\displaystyle A=\mathbb{R}$.

Similarly the other way round.