# how can I prove formally that this set is not path connected

Let $\mathbb{R}^2$ with its usual topology, let $D$ the set of all the lines that pass through the origin, with rational slope. And add to $D$ some point that does not lie in any of the lines ( call this new set $X$). For example $(e,e 2^{1/2})$ It is easy to show that this set is connected since, all the lines are connected and the intersection of these is not empty, then the union is also connected. And any set that lies between $D$ and its closure will also be connected, the closure is all $\mathbb{R}^2$ so $X$ is also connected. The question is if this set it´s also path connected, but I don´t think so , only for geometrically reasons. But in general it´s difficult to me, that this assertion )=. If someone can help me )=

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"it's" is a contraction; it stands for "it is". It is not the possessive (the possessive is "its"), and it is not a synonym for "is". – Arturo Magidin Sep 11 '11 at 19:51
sorry I do not speak english )= – Daniel Sep 11 '11 at 19:53
@Daniel: And I'm giving you a tip on how to write it better in the future. – Arturo Magidin Sep 11 '11 at 19:57
Hint: Argue by contradiction. Suppose there is a path connecting the new point $z$ to some point of $D$. Call the path $(x(t), y(t))$ for $0 \le t \le 1$. What can you say about the slopes $y(t)/x(t)$? – Nate Eldredge Sep 11 '11 at 19:58

Let $f$ be a path starting at the new point, with coordinate functions $f_1$ and $f_2$. Consider the function $\frac{f_2}{f_1}$, which starts out irrational, and must become rational if the path ever reaches an element of $D$. Between any irrational and rational number, there is an irrational number, and you can apply the intermediate value theorem to $\frac{f_2}{f_1}$.