The book I am using for my Introduction of Topology course is Principles of Topology by Fred H. Croom.

Prove that $A = (\mathbb{Q}\times\mathbb{R})\cup (\mathbb{R}\times\mathbb{Q})$ is a locally connected and connected subspace of $\mathbb{R}^2$

This is what I understand:

  • A space $X$ is connected provided that it cannot be written as the disjoint union of specifically two open sets.
  • A space $X$ is locally connected at a point $a$ in $X$ if every open set containing $a$ contains a connected open set which contains $a$. The space $X$ is locally connected provided that it is locally connected at each point.
  • I'm aware that neither connected nor locally connected implies the other, nor do their negations. Meaning I would have to prove each.
  • Now the set of rationals is disconnected (in fact totally disconnected). $\mathbb{R}^2$ is connected and locally connected.
  • $A \subseteq \mathbb{R}^2$

I'm a little stuck on how to approach this question; however, I do have a rough idea.

To prove $A$ is connected, I can claim that $A$ is disconnected. Then there exists two nonempty set $U$ and $V$ such that $U\cup V=A$ and $U \cap V = \emptyset$. Thus $U$ and $V$ are clopen. Eventually, I should arrive at a contradiction because $\emptyset$ and $A$ are the only subsets of $A$ that are clopen, proving $A$ is connected. However I am not sure how to properly execute the approach.

To prove $A$ is locally connected. I can claim the space $A$ has a local basis $\mathscr{B}_a$. I would need to prove that $\mathscr{B}_a$ consist of connected open sets, proving $A$ is locally connected.

Am I on the right track? Any suggestions oh how I can proceed with my ideas?

Sorry for the long read. If there are any mistakes in what I stated above, please let me know so I can correct it. I sincerely thank you for taking the time to read this question. I greatly appreciate any assistance you may provide.

  • $\begingroup$ I suspect that the space in question is supposed to be $(\mathbb{Q}\times\mathbb{R})\cup (\mathbb{R}\times\mathbb{Q})$. Please confirm, Kevin. $\endgroup$ Apr 6, 2015 at 18:32
  • $\begingroup$ @JohnHughes, yes that is what I meant. I corrected the mistake now. Thank you. $\endgroup$
    – Kevin_H
    Apr 6, 2015 at 18:34
  • 1
    $\begingroup$ Instead of your indirect approach, it's probably to prove directly that $A\cap(a,b)\times(c,d)$ is always path-connected. This then easily implies that $A$ is both connected and locally connected. $\endgroup$ Apr 6, 2015 at 18:38

2 Answers 2


For "connected", I suggest you show that it's path-connected. To find a path from $(q, r)$ to $(q', r')$, move along the edge $([q, 0], r)$ that's in the first component, and then along $(0, [r, r'])$ and then along $([0, q'], r')$. Once you make sense of this (looking at cases where $q < 0, q> 0, q = 0$, etc.), the rest should be straightforward.

In fact, just write down a path from $(q, r)$ to $(0, 0)$, and you're done (at least with this part). That should also help you do the second part. Why? Because there are a lot of rationals -- 0 isn't the only one.


There are several possible approaches to this. As suggested by John Hughes, you could prove path-connectedness. Another approach is as follows: $A$ is the union of $$A_q=\{q\}\times\mathbb R\cup\mathbb R\times\{q\}$$ over all rationals $q\in\mathbb Q$. Now, remember the following:

Theorem. Suppose $(X_\lambda)_{\lambda\in\Lambda}$ is a family of connected sets such that $X_{\lambda_1}\cap X_{\lambda_2}$ is non-empty for all $\lambda_1,\lambda_2$. Then $\bigcup_{\lambda\in\Lambda}X_\lambda$ is connected.

It follows easily from this that $A_q$ are connected and that $\bigcup_{q\in\mathbb Q} A_q$ is connected.

Your can prove local connectedness using a similar approach: first observe that the neighborhood of any point contains a "square", then prove that this "square" is connected.

  • $\begingroup$ Let us suppose that we choose the neighborhood around the origin $(0,0)$. Then what "square" would this look like and how would we go about showing it is connected? $\endgroup$
    – Jamil_V
    Apr 6, 2015 at 22:26
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    $\begingroup$ @Jamil_V: Let $Q_\epsilon=(-\epsilon,\epsilon)\cap\mathbb Q$ and $R_\epsilon=(-\epsilon,\epsilon)\cap\mathbb R$. Then the "square" would be $S=Q_\epsilon\times R_\epsilon\cup R_\epsilon\times Q_\epsilon$ for some $\epsilon$. We can then prove that it is connected using exactly the same argument as for $A$ except that $S$ is the union of $S_q=\{q\}\times R_\epsilon\cup R_\epsilon\times\{q\}$ over $q\in Q_\epsilon$. (For neighborhoods around other points, suitably translate everything.) $\endgroup$
    – Dejan Govc
    Apr 6, 2015 at 22:41

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