Given that A is a non-empty subset of real numbers, if I(A) denotes the set of interior points of A; then I (A) is:-

a) empty. b) singleton. c) a finite set containing more than one element. d) countable but not finite.

I know that the largest open set contained in A is called interior of A. Also a point is said to be interior point of A if we have an open ball of finite radius contained in A.

My trouble is I am not able to figure out the interpretation from the given statement as it is not given anything else beside non empty subset.

  • $\begingroup$ None of these options has to hold. Take $A=\mathbb{R}$. The interior is all of $\mathbb{R}$, which is nonempty, not a singleton, infinite, and uncountable. $\endgroup$
    – kccu
    Jan 10, 2016 at 5:03
  • $\begingroup$ But can I say option d) as answer if I dont take A to be R? But again problem occurs on open intervals ! For ex. (0,1) is uncountable. $\endgroup$
    – Kavita
    Jan 10, 2016 at 5:13
  • $\begingroup$ No, take for example $A=[0,1]$. The interior $(0,1)$ is uncountable. In fact, if $A$ has nonempty interior, $I(A)$ contains an open interval and is thus uncountable. $\endgroup$
    – kccu
    Jan 10, 2016 at 5:15
  • 1
    $\begingroup$ Are you sure you don't mean "boundary points" and not "interior points"? $\endgroup$
    – user2469
    Jan 10, 2016 at 5:31

1 Answer 1


The interior of any subset $S$ of $\mathbb R$ is open in $\mathbb R$, so it is either empty or a countable union of open balls in $\mathbb R$.

(In fact, I could omit the words "either empty or" above, since the union could be taken over an empty indexing set, which produces an empty union.)


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