Determine whether the following spaces are connected, totally disconnected or neither

1) $\mathbb{R}$ with the cocountable topology

2) $\mathbb{R}$ with the lower limit topology

3) $A=\{(x,y) \in \mathbb{R}^2|\text{ x is rational or y is rational}\}$

4) $B=\{(x,y) \in \mathbb{R}^2|\text{ x is rational and y are rational}\}$

5) $C=\{(x,y) \in \mathbb{R}^2|\text{ x is rational or y is rational but not both}\}$

My attempt

1) Suppose A and B are disjoint nonempty open sets such that $\mathbb{R}=A\cup B$, then both A and B are uncountable. But $B \subset A^c$ and $A^c$ is countable which is a contradiction. So $\mathbb{R}$ is connected.

2) Consider $[0,1)$ is a proper subset of $\mathbb{R}$ which is open. Now the complement of $[0,1)$ is $(-\infty,0)\cup[1,\infty)$ which is also open in $\mathbb{R}$ thus $[0,1)$ is also closed. Since \mathbb{R} contains a nonempty proper subset that is both open and closed it is disconnected. It is not totally disconnected since (0,1) is a connected subset of $\mathbb{R}$

3) Let $(a_1,a_2),(b_1,b_2)\in A$.

Case 1: If $a_1$ and $b_2$ are rational. Then $(\{a_1\}$x$\mathbb{R})\cup(\mathbb{R}$x$\{b_2\})$ is connected since each set in the union is connected and contain $(a_1,b_2)$. Thus $(a_1,a_2)$ and $(b_1,b_2)$ lie in a connected subset of A

Case 2: If $a_1$ and $b_1$ are rational. Then $(\{a_1\}$x$\mathbb{R})\cup(\mathbb{R}$x$\{0\})$ is connected since both contained $(a_1,0)$. Now $[(\{a_1\}$x$\mathbb{R})\cup(\mathbb{R}$x$\{0\})]\cup(\{b_1\}$x$\mathbb{R})$ is connected since both contain $(b_1,0)$. Thus $(a_1,a_2)$ and $(b_1,b_2)$ lie in a connected subset of A

Hence A is connected.

4) Since $B=\mathbb{Q}$x$\mathbb{Q}$ and $\mathbb{Q}$ is totally disconnected. It follows that B is totally disconnected because it is a product of totally disconnected space.

I'm not sure with (5) it seems to be disconnected but I'm not sure how to show it.

Can anyone help with number (5) and please check my argument for (1) to (4)

Thank you.


Your answers for (1), (3), and (4) are fine.

Your argument that (2) is not connected is correct, but in fact (2) is totally disconnected: for any $x,y\in\Bbb R$ with $x<y$ the sets $(\leftarrow,y)$ and $[y,\to)$ are disjoint clopen sets separating $x$ and $y$. Note that the interval $(0,1)$ is not connected in the lower-limit topology: it is the union of the disjoint open sets $\left(0,\frac12\right)$ and $\left[\frac12,1\right)$.

For (5) let $\Bbb P=\Bbb R\setminus\Bbb Q$, the set of irrational numbers; then $C=(\Bbb Q\times\Bbb P)\cup(\Bbb P\times\Bbb Q)$. For $r\in\Bbb Q$ let

$$U_r=\{\langle x,y\rangle\in C:x+y>r\}$$


$$V_r=\{\langle x,y\rangle\in C:x+y<r\}\;.$$

Note that $x+y\in\Bbb P$ whenever $\langle x,y\rangle\in C$, so $C=U_r\cup V_r$. Clearly $U_r$ and $V_r$ are disjoint and open in $C$, so $C$ is not connected. Moreover, points $p_0=\langle x_0,y_0\rangle$ and $p_1=\langle x_1,y_1\rangle$ in $C$ can be separated by such a pair of clopen sets whenever $x_0+y_0\ne x_1+y_1$. In fact we can also separate $p_0$ and $p_1$ when $x_0+y_0=x_1+y_1$, by cutting $C$ parallel to the line $y=x$ instead of parallel to the line $y=-x$, so $C$ is totally disconnected; I’ll leave it to you to find and describe that separation. You’ll want the fact that $\Bbb Q+r=\Bbb Q$ and $\Bbb P+r=\Bbb P$ whenever $r\in\Bbb Q$.

  • $\begingroup$ Thank you. Just to be sure, so if $x_0+y_0=x_1+y_1$, then it must be the case that $x_0=y_1$ and $x_1=y_0$, so our points is of the form (x,y) and (y,x) where either x or y is rational but not both. So I can instead take WLOG that x-y < r < y-x for some rational r. So $U_r = \{(x,y) \in C : x-y < r\}$ and $V_r= \{(x,y)\in C : x-y > r\}$ are the desired separation. Right?. $\endgroup$ – Sai Nov 10 '16 at 11:09
  • $\begingroup$ @Sai: No, if $x_0+y_0=x_1+y_1$ you might have $x_0=0,y_0=\pi,x_1=-1$, and $y_1=1+\pi$, for instance. All you know is that (in that particular case) both points are on the line $x+y=\pi$. But you do know that $x_0-y_0\ne x_1-y_1$, because otherwise $\langle x_0,y_0\rangle=\langle x_1,y_1\rangle$. Thus, there is indeed a rational $r$ between $x_0-y_0$ and $x_1-y_1$, and you can use it as you suggest to separate the points; the line $x-y=r$ is disjoint from $C$ and splits it into two clopen sets, one containing each of the points. (In fact $C$ has a base of clopen diamonds.) $\endgroup$ – Brian M. Scott Nov 10 '16 at 17:28
  • $\begingroup$ That was careless of me. Thank you for pointing that out. $\endgroup$ – Sai Nov 11 '16 at 3:57
  • $\begingroup$ @Sai: You’re welcome. $\endgroup$ – Brian M. Scott Nov 11 '16 at 18:46

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