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  1. Let S be a subset of Rn, and suppose that S is not bounded. Construct an open cover of S with no finite subcover, then prove this claim about your open cover.

  2. Let S be a subset of Rn such that S is not closed. Construct an open cover of S with no finite subcover, then prove this claim about your open cover.

I construct the open cover {Un} of S by constructing a sequence Pn in S with no subsequences converging to a point in U, and each Pi is in Ui, then by definition of compactness, this open cover has no finite subcover. I wonder if there are other ways to construct the open covers more directly.

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2 Answers 2

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For the first part, you can use an open cover $\{B_{N}(0)\}$ of balls of increasing size. Since the set is unbounded, there is no finite subcover.

For the second part, let $x \in \overline{S}\setminus S$. You can then use the open cover $\{ \Bbb R^N \setminus \overline{B_{1/n}(x)}\}$ of complements of closed balls. This is a cover since $x\not \in S$, so that any point $s\in S$ will lie in $\Bbb R^N \setminus \overline{B_{1/n}(x)}$ as long as $1/n < d(x,s)$. Since $x$ is a limit point of $S$ there is no finite subcover, as points of $S$ can be arbitrarily close to $x$.

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For the first part, consider the covering

$$\bigcup _{n=1}^{\infty} D_n (0)$$

(here, $D_n (0)$ denotes the open ball of radius $n$ centered at $0$).

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