# Let $S \subset \mathbb{R^2}$ consisting of all points $(x,y)$ in the unit square $[0,1] × [0,1]$ for which $x$ or $y$, or both, are irrational.

Full question: Let $$S$$ be the subset of $$\mathbb{R^2}$$ consisting of all points $$(x,y)$$ in the unit square $$[0,1] × [0,1]$$ for which $$x$$ or $$y$$, or both, are irrational. With respect to the standard topology on $$\mathbb{R^2}$$, $$S$$ is:

A - closed, B - open, C - connected, D - totally disconnected, E - compact.

So if $$x$$ is irrational and $$y$$ is rational, we just have the unit square and I'd say thats $$A,C$$, and $$E$$. And if both are irrational, I'd say its $$D$$ because the set of $$\mathbb{I}^2$$ is totally disconnected. However the answer key says the answer is $$C$$.

Can someone help explain?

It is path-connected, and we can simply find a path between any two points in S. For example, if $$a$$ is rational and $$b,c,d$$ are irrational, we construct a path $$(a,b) \to (c,d)$$ as follows. First pick some irrational number $$r$$, then go $$(a,b) \mapsto (r,b) \mapsto (r,d) \mapsto (c,d)$$, taking straight lines between the mentioned points. Other cases have similar paths. The result then follows since path-connected $$\implies$$ connected.
• If you have $a,b,c,d$ all irrational, then you can just trace straight lines $(a,b) \mapsto (c,b) \mapsto (c,d)$, and those straight lines are each in $S$. May 27 '20 at 20:44
• @TopologicalGeomiter since $b$ is irrational, the point $(x,b)$ is in $S$ for any $x \in [0,1]$, so the straight line from $(a,b)$ to $(c,b)$ is contained in $S$. May 29 '20 at 21:07
• @TopologicalGeomiter rereading your post, I think we've interpreted the question differently. My interpretation was $S = \{ (x,y) \in [0,1]^2 : \text{ at least one of } x, y \text{ is irrational} \} = [0,1]^2 \cap (\mathbb{R}^2 \setminus \mathbb{Q}^2)$. May 29 '20 at 21:09