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Question: What is the closure of the set $\{(-1/2)^n \, : \, n \in \mathbb{N} \} \cup \{0\}$ as a subset of $\mathbb{R}$? Is it compact? Is it connected?

My answer: The closure is the set itself. It is compact as the set is clearly bounded in $\mathbb{R}$ and closed as the closure is itself. No it's not connected in $\mathbb{R}$ as any two open intervals in $\mathbb{R}$ where the union is the subset is a counterexample.

Just checking I'm right?

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The last bit seems vague to me... Specify those open sets. Also, you should provide an argument as to why the set is closed (I would use sequences). –  David Mitra Mar 4 '12 at 18:33
You need to be careful with your proof of connectedness. First of all, the set you describe is not the union of any two open intervals of $\mathbb{R}$; you probably meant to pick two open intervals which cover the set. Secondly, if it not enough to take any two such intervals; note that $(-2,\frac{1}{2}) , (\frac{-1}{2},2)$ are open intervals which cover the set, but these will not witness the disconnectedness of the set in question. (You need to ensure that these open intervals do not meet in the set in question.) –  Arthur Fischer Mar 4 '12 at 19:09

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Denote your set by $A$. If we show that the closure of $A$ is itself, then we will have shown that $A$ is closed. One way to show that a set is closed is to show that, for any $p \notin A$, there is an open set $U$ containing $p$ such that $A \cap U = \emptyset$ (i.e. we are showing $A^c$ is open). Since there are "large" gaps between any points of $A$, you can always get an open interval around any point of $A^c$ that misses points of $A$. I'll leave the details to you.

$A$ is compact for the reason you give (the Heine-Borel Theorem).

$A$ is not connected. For a concrete example, let $U$ be the open interval $(.9, 1.1)$ and $V$ be the open interval $(-1, .9)$. The interval $U$ contains only one point of $A$ (namely, 1), while $V$ contains all the rest. (In fact, this space would be totally disconnected if you removed $\{0\}$.)

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