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Let $A$ be a closed subset of $\mathbb R$, satisfying $A \neq \emptyset$ and $A \neq \mathbb R$. Then $A$ is

  1. the closure of the interior of $A$.

  2. a countable set.

  3. a compact set.

  4. not open.

Argument:

Since, $A$ is subset of $\mathbb R$, it should be an interval or singleton set.

$A$ cannot be countable. counter example is $[0,1]$ so (2) is not the answer.

$A \neq cl(int(A))$. counter example any singleton set $A={1}$. $int(A) = \phi$

$cl(int(A)) = \phi \neq A$.(Hence (1) is not the answer)

Definitely $A$ is not open (whole set and empty set are only sets which are both open and closed) but what about Compactness, Since $A$ is already closed, if it is bounded, then it is compact. Since any subset of $\mathbb R$ is of $[-a,b]$ which is bounded, hence $A$ is compact and $A$ is also not open.

Am i right? or please indicate where my argument is wrong..

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  • $\begingroup$ A subset of $\mathbb{R}$ does not have to be an interval or a singleton. What about $A=\{1\}\cup[2,3]$? $\endgroup$ – parsiad Apr 12 '16 at 12:42
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Your wording is confusing. For example, you use "cannot" where you mean "is not necessarily:"

  1. $A$ is not necessarily the closure of its interior; consider $A=\{0\}$.

  2. $A$ is not necessarily countable; consider $A=[0,1]$.

  3. $A$ is not necessarily compact; consider $A=\mathbb{N}$.

  4. Suppose $A$ is open. Since $A\neq\emptyset$ and $A\neq\mathbb{R}$, it follows that $\mathbb{R}$ is not a connected space; a contradiction. Therefore, $A$ is not open.

I would work on your mathematical writing. Compare the above with your answers and see where they differ.

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  • $\begingroup$ Thanks for the answer. sure i will work on my mathematical writing. $\endgroup$ – Sam Christopher Apr 12 '16 at 13:06
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  1. False, take a singleton for instance

  2. Not necessarily, take any closed interval.

  3. False, take $[1,+\infty[$ for instance

  4. True, since $\mathbb R$, the only subset that is both closed and open is $\mathbb R$ itself.


As pointed out in the comments, you are wrong to assert that A can only be a singleton or an interval, it can also be any finite union of both.

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    $\begingroup$ Yes not necessarily. But we are sure by definition that any finite union of closed subset is itself closed. That was a non-exclusive sentence ;) $\endgroup$ – pmichel31415 Apr 12 '16 at 12:54

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