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Prove that $X=\{(x,y) :x\text{ is rational or }y\text{ is rational}\}$ is path connected.

So for every $(x,y)$ in $X$, I need to find a continuous function $f$ on $[a,b]$ such that $f(a)=x$ and $f(b)=y$.

Trying to think about suitable functions, maybe one which sends everything to nearest rational. But don't think such a function exists.

user193322
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3 Answers3

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Indeed, a "nearest rational" does not exist. If $x$ is rational, you can find an obvious path from $(x,y)$ to $(x,0)$ and then on to $(0,0)$. Similarly, if $y$ is rational, you can find a path to $(0,0)$ via $(0,y)$. Thus any $(x,y)\in X$ is pathconnected to $(0,0)$.

Hagen von Eitzen
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Proof guideline:

  1. Show that for all $ x \in \mathbb{Q}$ any line $L_x = \{(x,y) , \forall y \in \mathbb{R}\}$ is path-connected
  2. Show that for all $ y \in \mathbb{Q}$ any line $L'_y = \{(x,y) , \forall x \in \mathbb{R}\}$ is path-connected
  3. From (1) and (2) follows that $L_0$ and $L'_0$ are path-connected
  4. Show that if $X$ and $Y$ are path connected and $X \cap Y \neq \emptyset$, then $X \cup Y$ is path connected

Explanation:

Notice that if you would "draw" $X$ it would look like a grid of perpendicular lines. Now each of these lines cross either the x-axis or the y-axis. Since $0$ is a rational number, both the x-axis and the y-axis are subsets of $X$. Each line is now connected to another through the axis. So by proving point (4) we show that the whole of $X$ is also path-connected.

d1gits
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Path connected means that for arbitrary pairs of points in $X$, $(x_1, y_1)$ and $(x_2, y_2)$, you can find a continuous map $f : [0,1] \rightarrow X$ such that $f(0) = (x_1, y_1)$ and $f(1) = (x_2, y_2)$.

One way to do this is to construct a path $(x_1, y_1)$ to $(x_1,0)$ to $(0, 0)$ to $(x_2, 0)$ to $(x_2, y_2)$. That is, every member of $X$ is path connected with $(0,0)$.

Simon S
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  • I'm a bit confused. So you're saying use the composition? Composition of cont. functions is continuous. How do I know function is continuous? – user193322 Nov 24 '14 at 13:22
  • I've edited to answer to remove the $\rightarrow$ and replaced them with "to". The path goes point to point to point; not a composition of functions. – Simon S Nov 24 '14 at 13:25
  • I still don't get what function I'm meant to be using. How do you get from a path to a function. – user193322 Nov 24 '14 at 13:40
  • For example a path $p_1 : [0, 1] \rightarrow X$ from $(x_1, y_1)$ to $(x_1, 0)$ would be given by $f(t) = (x_1, (1-t)y_1)$. Note that $f(0) = (x_1, y_1)$ and $f(1) = (x_1, 0)$. – Simon S Nov 24 '14 at 13:44
  • Hi @SimonS .. sorry bother you, but to avoid create new topics, can you answer me a dumb question about this problem ? $p_1$ being continuous, for example the second coordenate rational, ie, $\mathbb R$X$ \mathbb Q$, inverse image open don't need to be open in $[0,1]$? but theres not open sets in $\mathbb Q$ .. i am little confused.. Thank you very much. – Francisco May 07 '15 at 22:56
  • No subset of $\mathbb Q$ is open in $\mathbb R$ with the latters usual topology. But if $\mathbb Q$ takes the induced topology from $\mathbb R$, then there are lots of open sets in $\mathbb Q \ $! – Simon S May 07 '15 at 22:58