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Consider $\mathbb{R}$ with the usual topology and $\mathbb{R}^{2}$ with the product topology. Show that the set, $$X=\{ (x,y)\mid 0\leq x\leq 1, x\in\mathbb{Q}, 0\leq y\leq 1 \}\cup \{ (x,y)\mid 0\leq x\leq 1, x\in\mathbb{I}, -1\leq y\leq 0 \}$$ is connected. Edit: $\mathbb{I}$ is the set of irrationals numbers

My approach: If I can show that, $X$ is a linear continuum in the order topology, then $X$ is connected. Where a simply ordered set $X$ having more than one element is called a linear continuum, if the following hold:

$X$ has the least upper bound property. And, if $x<y$, there exist $z$ such that $x<z<y$.

But, I though prove that $X$ is path connected, and then $X$ is connected. But in this case, How can construct a path for every point in $X$, I think the pasting Lemma is very useful, but I dont see how. Thanks!

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  • $\begingroup$ What's $\Bbb I$? Is it $\Bbb R\setminus\Bbb Q$? $\endgroup$ – user228113 May 30 '17 at 7:23
  • $\begingroup$ My guess is that $\mathbb{I}=[0,1]$. $\endgroup$ – José Carlos Santos May 30 '17 at 7:27
  • $\begingroup$ @JoséCarlosSantos That would make little sense in this case, if you look at where it is used. $\endgroup$ – user228113 May 30 '17 at 7:28
  • $\begingroup$ @G. Sassatelli You are right. I wasn't paying attention. $\endgroup$ – José Carlos Santos May 30 '17 at 7:29
  • $\begingroup$ There is nothing linear about X. $\endgroup$ – William Elliot May 30 '17 at 8:21
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What is $\mathbb{I}$? Well, no matter what it is, as long as $$\big([0,1]\cap\mathbb{Q}\big)\cup\big([0,1]\cap\mathbb{I}\big)=[0,1]$$

then the statement is true. That's because in this situation $[0,1]\times\{0\}\subset X$.

Let's write

$$A=\{(x,y)\ ∣\ 0\leq x\leq 1, x\in\mathbb{Q}, 0\leq y\leq 1\}$$ $$B=\{(x,y)\ ∣\ 0\leq x\leq 1, x\in\mathbb{I}, -1\leq y\leq 0\}$$ $$X=A\cup B$$

Pick a point $(x,y)\in A$. Then there is a vertical path

$$\lambda:[0,1]\to X$$ $$\lambda(t)=(x, ty)$$

Thus we've connected $(x,y)$ with $(x, 0)$. The same argument shows that we can connect any point $(x,y)\in B$ with $(x,0)$. Now since $[0,1]\times\{0\}\subset X$ then we can also connect two points $(x_1, 0)$ and $(x_2, 0)$ via

$$\lambda:[0,1]\to X$$ $$\lambda(t)=\big(tx_1+(1-t)x_2, 0\big)$$

This shows that any two points can be connected via a path (a composition of at most 3 paths defined above) and thus $X$ is path connected.


Conversly, if

$$\big([0,1]\cap\mathbb{Q}\big)\cup\big([0,1]\cap\mathbb{I}\big)\neq[0,1]$$

then the statement is not true. That's because we have a continous projection

$$\pi:X\to[0,1]$$ $$\pi(x,y)=x$$

Now pick $$a\in[0,1]\backslash\bigg(\big([0,1]\cap\mathbb{Q}\big)\cup\big([0,1]\cap\mathbb{I}\big)\bigg)$$

and define $U=[0,a)$ and $V=(a, 1]$ and note that

$$X=\pi^{-1}(U)\cup\pi^{-1}(V)$$

And both sets on the right side are nonempty and open. Thus $X$ is not connected.

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  • $\begingroup$ any way, $\mathbb{I}$ is the set of irrational numbers. $\endgroup$ – user400623 May 30 '17 at 18:54
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I'll answer your question ignoring your $x\in\mathbb{I}$ condition.

Let $A(x):=\{x\}\times [0,1]$, and let $B:=[0,1]\times[-1,0]$. Clearly, $C(x):=A(x)\cup B$ is connected for every $x\in[0,1]$, and in particular, it is connected for every such rational $x$. Also clearly $C(x)\cap C(x')=B\neq\emptyset$ for all $x\neq x'$. So your set is the union of connected sets $C(x)$ that have non-empty intersection. Consequently your set is connected.

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