the question I'm having trouble with is this:

$$A:=\bigcup_{n=1}^{\infty}\left({\left[\frac{1}{n+1},\frac{1}{n}\right) \times \left(0,n\right)}\right).$$

In the Euclidean space $\mathbb{R}^2$, is the subset $A$ either closed or open?

I drew a picture in $\mathbb{R}^2$ and I really think it's open, but I can't come up with an elaborate way to describe why. Any help would be appreciated.


  • $\begingroup$ The two intervals are both open in $\mathbb{R}$, you can use this. $\endgroup$ – Ninja Apr 19 '17 at 12:19
  • $\begingroup$ how is the first one open? $\endgroup$ – DHMO Apr 19 '17 at 12:20
  • $\begingroup$ First one is $(0,1)$, is not it? $\endgroup$ – Ninja Apr 19 '17 at 12:20
  • 1
    $\begingroup$ but union of products is not product of unions $\endgroup$ – DHMO Apr 19 '17 at 12:21
  • $\begingroup$ adding parentheses... $\endgroup$ – GEdgar Apr 19 '17 at 12:26

It's open. To see this, note that the only problematic points are $(\frac1{n+1}, y)$ for $y \in ]0,n[$. But $]\frac1{n+2}, \frac1{n}[ \times ]y-\epsilon, y+\epsilon[ \subset$ of the original set, where $\epsilon = \min\left\{ \frac{n-y}{2}, \frac{y}{2} \right\}$.

  • $\begingroup$ Your name perfectly fits in this question. $\endgroup$ – DHMO Apr 19 '17 at 12:30
  • $\begingroup$ @DHMO I've had better ones. $\endgroup$ – user384138 Apr 19 '17 at 12:31
  • $\begingroup$ Side question about nomenclature: is $(-{\infty},0] \bigcup \space [n,{\infty})$ the same as $]0,n[$ ? I haven't seen the backwards brackets before. $\endgroup$ – Burnsba Apr 19 '17 at 14:00
  • $\begingroup$ @Burnsba Your two sets are complements. $\endgroup$ – G. Bach Apr 19 '17 at 15:52
  • $\begingroup$ Thanks for answering! But I noticed I had one more question to solve. Would the closure of this set A be $\bigcup_{n=1}^{\infty}\left(\;{\left[\frac{1}{n+1},\frac{1}{n}\right] \times \left[0,n\right]}\;\right)$ then? I used $cl(A)=A\bigcup acc(A)$, and thought that the boundaries would be included in acc(A) $\endgroup$ – Peter Apr 19 '17 at 16:08

Pick a point in the set. It must belong to one of the products in the union.

Say, it belongs to $\left[\frac{1}{k+1},\frac{1}{k}\right[ \times \left]0,k\right[$.

Let the point be $(x,y)$. We have $\frac1{k+1} \le x < \frac1k$ and $0 < y < k$.

Now, we consider $\left]\frac1{k+2},\frac{x+1/k}2\right[ \times \left]\frac y2,\frac{y+k}2\right[$ and prove that it is contained in $A$.

The set is also open

Since there is an open set around every point in $A$, we can conclude that $A$ is open.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.