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, 2017 at 12:19
  • $\begingroup$ how is the first one open? $\endgroup$
    – DHMO
    Apr 19, 2017 at 12:20
  • $\begingroup$ First one is $(0,1)$, is not it? $\endgroup$
    – Ninja
    Apr 19, 2017 at 12:20
  • 1
    $\begingroup$ but union of products is not product of unions $\endgroup$
    – DHMO
    Apr 19, 2017 at 12:21
  • $\begingroup$ adding parentheses... $\endgroup$
    – GEdgar
    Apr 19, 2017 at 12:26

2 Answers 2


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, 2017 at 12:30
  • $\begingroup$ @DHMO I've had better ones. $\endgroup$
    – user384138
    Apr 19, 2017 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, 2017 at 14:00
  • $\begingroup$ @Burnsba Your two sets are complements. $\endgroup$
    – G. Bach
    Apr 19, 2017 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, 2017 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.


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