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Please help me to prove or disprove the following:

1.$\text{}$ If $\{F_i\mid i=1,2,\ldots\}$ is a collection of closed sets in $\mathbb{R}^2$, then the union of the $F_i$ is also closed in $\mathbb{R}^2$.

By the properties of closed sets, we can say that finite union of closed sets is closed. But how can we construct a counterexample to show that the arbitrary union is not closed in $\mathbb{R}^2$?

2.$\text{}$ In a topological space, the intersection of $\mathrm{int}(A)$ with the derived set of $A$ ($A^d$) is empty.


3.$\text{}$ $\{1-\frac{1}{n}\mid n=1,2,\ldots\}$ is closed in $\mathbb{R}$.

A set is closed iff it contains all of its limit points. Is zero a limit point of this? If so this set is closed. Am I correct?

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2 Answers

up vote 1 down vote accepted

For part 1, we know that $\mathbb{Q}$ is a countable subset of $\mathbb{R}$, which is not closed (because every point of $\mathbb{R}$ is a limit point of $\mathbb{Q}$).

$\mathbb{R}$ is open, so $int(\mathbb{R}) = \mathbb{R}$. However, every point of $\mathbb{R}$ is a limit point, thus $int(\mathbb{R})\cap\mathbb{R}^d = \mathbb{R}$.

Take the point 1. Clearly it is not an element of our set (call it $S$). Then, given any $\varepsilon > 0$, we may find $N(\varepsilon) \in \mathbb{N}$ such that $a_n = 1 - \frac{1}{n} \in B(1,\varepsilon)$ whenever $n > N(\varepsilon)$. Thus $S$ does not contain all of its limit points, and is not closed.

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@ Andy,Thank you very much Andy,these answers are very clear! –  ccc Dec 26 '12 at 5:25
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1) Maybe you can work out a way to write 3 as a union of closed subspaces?

2) What's the interior of $\mathbb R$. What's the derived set?

3) Is $1$ a limit point of the set? Is it in the set?

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@ jacob,thank you very much! –  ccc Dec 26 '12 at 5:24
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