I am working on the following question with the question specific definition that: a set $A\subset\mathbb{R}$ is nowhere dense iff for any two points $c<d,\exists a<b\ s.t.\ [a,b]\subset[c,d]$ and $[a,b]\cap S=\varnothing.$

If A is countable, show that A is a union of countably many nowhere dense sets(Hint: Singleton).

I know that countable union of countable sets are countable but I havent a clue how to go about proving a singleton is nowhere dense using the supplied definition alone. I get that interior of a closure must be empty also, but not given that, and not sure how to derive that from given definition.

My current argument going with that A is a singleton by being countable, $$ [a,b]\cap A=\varnothing\Rightarrow([a,b]\cap A)^\circ=(\varnothing)^\circ\Rightarrow[a,b]^\circ\cap A^\circ=\varnothing\Rightarrow(a,b)\cap \varnothing=\varnothing. $$

Is this enough?

  • 11
    $\begingroup$ Did you read the hint? Do you know what a singleton is? Do you see that a singleton is nowhere dense? $\endgroup$ Apr 10, 2018 at 23:06
  • $\begingroup$ My point is that I was struggling to prove that a singleton was nowhere dense using the definition alone. $\endgroup$
    – cemsicles
    Apr 11, 2018 at 22:12

1 Answer 1


Let $A = \{a_n\}_{n = 1}^\infty$. Then we may also write $A = \bigcup_{n = 1}^\infty \{a_n\}$. Is this union countable? It suffices to show that if $x \in \mathbb{R}$, then the singleton $\{x\}$ is nowhere dense.

Why is this true? A set $A$ is nowhere dense if in every interval $(a, b)$ in $\mathbb{R}$, we can find some subinterval $(c, d)$ such that NO points of $A$ are in $(c, d)$. Contrast this with the definition of density; a set $A$ is dense if given any interval, there is always a point of $A$ within this interval.

How to prove that a one-point set is nowhere dense? Let $(a, b)$ be an interval in $\mathbb{R}$. If $x \leq a$ or $x \geq b$, we are done (since then $x$ is not in $(a, b)$ at all). So suppose $a < x < b$. Then the interval $(a, x)$ is contained in $(a, b)$, but $x \not\in (a, x)$.

  • $\begingroup$ Thanks for the reply. If the set is countable, then I know that countable union of countable sets is also countable. My point is I am not entirely sure how to show if each set is or a countable set for that matter is nowhere dense using that definition alone. My brain is melted at this point and I feel like I am missing something. $\endgroup$
    – cemsicles
    Apr 10, 2018 at 23:15
  • 2
    $\begingroup$ No problem! I'll edit my answer to address this. $\endgroup$
    – Chris
    Apr 10, 2018 at 23:17
  • $\begingroup$ I just noticed that you are using open intervals while the definition uses closed intervals. Could you clarify why this is the case? And also I thought the point was to show the intersection of $[c,d]\cap A=\varnothing$ (from your notation) but you have shown [a,b]. thank you $\endgroup$
    – cemsicles
    Apr 11, 2018 at 22:07
  • 3
    $\begingroup$ Density is typically defined in terms of open intervals, not closed intervals. In fact, with the definition you gave, singletons wouldn't be nowhere dense, since the closed interval $[x, x]$ does not contain a subinterval which does not intersect $x$. $\endgroup$
    – Chris
    Apr 11, 2018 at 22:50
  • 1
    $\begingroup$ I see why I was struggling to come up with my own answer even after reading yours. I kept using closed intervals and singleton always remained within. $\endgroup$
    – cemsicles
    Apr 11, 2018 at 22:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .