Consider the following subset $X \subset \mathbb{R^2}$ with the Euclidian subspace topology.

$\begin{multline} X=\{(0,y): y \in [-1,1]\} \cup \\ \{(x,y): x\in (0,1) , y \in \mathbb{Q} \cap[-1,1]\}\\ \cup \{(x,y) \in \mathbb{R^2}: x \in (-1,0), y \in (\mathbb{R \setminus Q)} \cap[-1,1]\}\text{.}\end{multline}$

(i) Is $X$ connected?

(ii) Find closure and interior of $X$

I tried to visualize the open sets in the subspace topology. They are intersections of "euclidian balls" with the set $X$. I think that the following set is not connected because if I call $A=\{(0,y): y \in [-1,1]\} \cup \{0\}\times [-1,1]$ and $B=\{(x,y) \in \mathbb{R^2}: x \in (-1,0), y \in (\mathbb{R \setminus Q)} \cap[-1,1]\}$ then I have, just from the definition of $X$:

$A \cup B= X$

$A \cap B=\emptyset$.

I think that $A$ and $B$ are open sets because they are the intersection of $X$ with a "Euclidian rectangle"...

Since $X$ is open, then $\operatorname{Int}(X)=\emptyset$ and $\operatorname{Cl}(X)=X$, but I'm not so sure of this.

  • $\begingroup$ One should always have $\operatorname{Int}(X) \subseteq X \subseteq \operatorname{Cl}(X)$. Maybe your confusing interiors and closures? $\endgroup$ – Henno Brandsma Jul 16 '17 at 23:30
  • $\begingroup$ thanks, I edit. I was copying from the paper $\endgroup$ – VoB Jul 16 '17 at 23:40
  • $\begingroup$ I cannot make grammatical sense of your def'n of A. In any case X is path-connected (and hence connected) because when (x,y) , (x',y') are in A , we can travel ( within X ) horizontally from (x,y) to (x,0) and vertically from (x,0) to (x',0) and horizontally from (x',0) to (x',y'). $\endgroup$ – DanielWainfleet Jul 20 '17 at 2:00

The space $X$ is arcwise connected, hence also connected. It's interior is empty, because $X$ doesn't contains open balls. On the other hand it's closure is the square $[-1,1]\times[-1,1]$, by the density of rational numbers in the set of real numbers.

  • 1
    $\begingroup$ ... and by the density of the irrationals in the reals $\endgroup$ – timon92 Jul 16 '17 at 22:04
  • $\begingroup$ thanks for the answer! I can't understand how you can say that $X$ doesn't contain open balls... And for the closure: how can I say that $[-1,1]\times[-1,1]$ is a closed set? Just because I got the subspace Euclidian topology ? $\endgroup$ – VoB Jul 16 '17 at 22:07
  • $\begingroup$ Note that open balls are convex subsets of the plane while $X$ doesn't contain vertical segments other than those contained in $\{0\}\times[-1,1]$; thus $X$ cannot contains open balls. $\endgroup$ – Fabio Lucchini Jul 16 '17 at 22:13
  • $\begingroup$ Recall that $[-1,1]$ is a closed subset of the real line, hence $[-1,1]\times[-1,1]$ is a closed subset of the plane. $\endgroup$ – Fabio Lucchini Jul 16 '17 at 22:14
  • $\begingroup$ Just a question: by the definition of $X$ the text gave me, I considered, for example ${(x,y):x∈(0,1),y∈ℚ∩[−1,1]}$ as a cartesian product. Is it right? Maybe I misunderstood the text. $\endgroup$ – VoB Jul 16 '17 at 22:21

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