I am struggling to work out a way of determining whether a set is connected or not. I have 3 examples I am looking it, and I am struggling to figure out which of them are connected/path-connected.

  1. $C=\{(q,y)\in\Bbb{R}^2:q\in\Bbb{Q},y\in[0,1]\}\cup(\Bbb{R}\times\{1\})$

I would say this is not connected, as we know that $\Bbb{Q}$ is not connected, and it is the union of an interval that I would say is not connected ($C=\{(q,y)\in\Bbb{R}^2:q\in\Bbb{Q},y\in[0,1]\}$ with a line-and I am not sure if this changes things?


  1. the set of all points in $\Bbb{R}^2$ with at least one coordinate in $\Bbb{Q}$

I would say this is not connected, as rational numbers converge to irrational numbers, and vice versa, but the irrational numbers are still in $\Bbb{R}$, so all the points are abritrarily close together? For example, we could fix one of the points as a rational number, and then see that the other coordinate could be all of $\Bbb{R}$ which is connected, so we cannot split this into a disjoint union of open sets, and thus we cannot split the whole of this set into a disjoint union? (This is very informal I know, was just my thoughts on it-although may be wrong.)

  1. $C\backslash(1,1)$, where C is as it was in 1.

As I couldn't work out whether $C$ was connected or not, I am quite stumped by this.

Any help appreciated, thanks.


Generally, showing something is not connected is easy in principle, you 'just' have to divide it into 2 non-empty open sets. Of course, finding those sets is the hard part, as usual having seen examples is the way to start and build an intuition for that.

Showing connectedness by the definition is hard, because it is often hard to prove that something with that many possibilities does not exist. Knowing that path-connectedness implies connectedness is a big help here, because showing that 2 arbitrary points can be connected by a path inside the set is often managable. The really hard cases are when something is not path conneced by still connected.

Problem 1) can be answered by first observing that each point $(q,y) \in \mathbb Q \times [0,1]$ is path connected to the point $(q,1)$, via

$$f: [0,1] \to C \text{ with } f(t)=(q,y+t(1-y)),$$

and two points on $\mathbb R \times \{1\}$ are obvioiusly path-connected to each other. Path connectedness is transitive, so any point in $C$ is path connected to any other via at most 2 points on $\mathbb R \times \{1\}$. So the $C$ in problem 1 is path connected and hence connected.

For problem 2) the argument is that a point with rational $x$-coordinate $(q,y)$ (via essentially the same kind of path function as in problem 1) is path connected to $(q,0)$ which is path connected to $(0,0)$. The same goes for a point with rational $y$-coordinate: $(x,q)$ is path connected to $(0,q)$ which is path connected to $(0,0)$. Again by transitivity of path connectedness, this makes the whole $C$ of problem 2 path connected and thus connected.

Problem 3) is IMO the hardest. Removel of $(1,1)$ from the $C$ of problem 1 makes the line $\mathbb R \times \{1\}$ become two rays instead: $(-\infty,1) \times \{1\}$ and $(1,\infty) \times \{1\}$, which are together not path connected.

Now it becomes important to recall the proof that path-connectedness implies connectedness. If a space is not connected, it can be partitioned into 2 open non-empty sets $A$ and $B$. If a path-connected space was not connected, one could find points $a \in A$ and $b \in B$, and then construct a path between them. It then turns out that this is not really possible.

Now assume that the $C$ from problem 3 is not connected, so it can be partitioned into 2 non-empty sets $A$ and $B$. If we can find path-connected 'commponents' in $C$, that means sets of points that are all path connected between themselves, the above proves that each such component must belong completely to either $A$ or $B$, you cannot have some points belonging to $A$ and others to $B$!

Using the same arguments as in problem one, one can now find that $C$ consists of 3 such path connected components:

$$C_1: \{(q,y) \in \mathbb Q \times [0,1]: q > 1\}\, \cup \, (1,\infty) \times \{1\}$$ $$C_2: \{(q,y) \in \mathbb Q \times [0,1]: q < 1\}\, \cup \, (-\infty,1) \times \{1\}$$ $$C_3: \{1\} \times [0,1)$$

We now have to consider only how these 3 components can be distributed among $A$ and $B$. Since neither $A$ and $B$ can't be empty, it means 2 components must go into one set, and one component is the sole other set.

Now it turns out that any ball around any point in $C_3$ will by necessity include points from $C_1$ (in the $\{(q,y) \in \mathbb Q \times [0,1]: q > 1\}$ part) and points from $C_2$ (in $\{(q,y) \in \mathbb Q \times [0,1]: q < 1\}$). That means $C_3$ must go into the same set as $C_1$ and $C_2$, which leaves the other set empty, which is forbidden.

So we finally see that although the $C$ of problem 3 is not path connected, but nonetheless connected.

  1. To find the path-connection between $(q_1, y_1)$ and $(q_2, y_2)$, move from the first point to $(q_1, 1)$, then from that point to $(q_2, 1)$ and lastly from that point to $(q_2, y_2)$. Path-connectedness then implies connectedness.

  2. Let $(q_1, y)$ and $(x, q_2)$ be two such points. A path between those two points are the line segments between the first point and $(q_1, q_2)$ and $(q_1, q_2)$ and the second point. The space is path-connected, hence connected. (Similar proofs if both x-coordinates or y-coordinates are rational).


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    $\begingroup$ What about the point $(\pi,1)$ in the first example? $\endgroup$ – Christoph Jun 14 at 19:30
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    $\begingroup$ Example to first is incorrect. Note that $\mathbb R \times \{1\}$ is part of the set, which makes your argument incorrect. $\endgroup$ – Ingix Jun 14 at 19:30
  • $\begingroup$ I completely missed the fact that $\mathbb{R} \times \{1\}$ is part of the set, so you are both correct. $\endgroup$ – Alexander Geldhof Jun 14 at 19:32
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    $\begingroup$ How is the space in the third example related to the argument in the second? It is the first example minus the point $(1,1)$. $\endgroup$ – Christoph Jun 14 at 19:33
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    $\begingroup$ @TedShifrin Yep, it has $3$ path components that consitute a single connected component :-) $\endgroup$ – Christoph Jun 14 at 19:36

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