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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?

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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.

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  • $\begingroup$ This makes sense if either x or y is rational, but what if both are irrational? $\endgroup$
    – user444540
    May 27 '20 at 16:47
  • $\begingroup$ 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 $S$. $\endgroup$ May 27 '20 at 20:44
  • $\begingroup$ How do we move from (a,b) to (c,b)? What I think is happening is we have R^2, and if x is rational and y is irrational, we don't run into any problems. If both are irrational, then the subset of R^2 is just this grid of points, impossible to get from one to another without needing a rational to bridge the gap. $\endgroup$
    – user444540
    May 28 '20 at 17:04
  • $\begingroup$ @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 $S$. $\endgroup$ May 29 '20 at 21:07
  • $\begingroup$ @TopologicalGeomiter rereading your post, I think we've interpreted the question differently. My interpretation was $S = \{ (x,y) \in [0,1]^2 : \text{ at least one of } x, y \text{ is irrational} \} = [0,1]^2 \cap (\mathbb{R}^2 \setminus \mathbb{Q}^2)$. $\endgroup$ May 29 '20 at 21:09

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