# Show that $T$(contains segments) is path connected

Consider dictionary order topology in $\mathbb{R}^2$.

Let $X$ denotes the rational points of the interval $[0,1]\times 0$ in $\mathbb{R}^2$. Let $T$ denotes the union of all line segments joining the point $p=0\times 1$ to points of $X$.

Show that $T$ is path connected.

Well, that is easy: Note that every line segments in $T$ are connected with $p$. Hence path connected.$\tag{**}$

But: Consider $X=[0,1]\times 0$ in $\mathbb{R}^2$ and $T$ denotes the union of all line segments joining the point $p=0\times 1$ to points of $X$.

In this case, I can use the same argument as before and claim that $T$ is path connected. But clearly $T$ is not path connected for the fact $[0,1]^2$ is not path connected.

So, is my argument (**) is vague? Please someone help!

• Why do you think the standard square is not path connected? – M. Van Sep 20 '17 at 14:10
• @M.Van It is shown in Munkres Topology that ordered square is connected but not path connected (Example 6, page 156). That is why I did not wrote that. – topology_001 Sep 20 '17 at 14:14
• Ah I see, not the standard topology on the square. – M. Van Sep 20 '17 at 14:28
• Could you give the topology on each of the topological spaces you talk about? – M. Van Sep 20 '17 at 14:30
• @M.Van it is dictionary order topology. Otherwise it would be mentioned in the question.(it is mentioned that if nothing written then dictionary order topology.) – topology_001 Sep 20 '17 at 14:34

As I understood, $T$ consists of the segments $[x, p]$, where $x\in [0,1]\times 0$. When $x\ne 0\times 0$, the open segment $(x, p)$ is a discrete subspace in the dictionary order topology, so this don’t imply path-connectedness of $T$. Also, something is wrong, because if you consider on $T$ the order topology induced from $\Bbb R^2$ then $T$ is clearly not path-connected, because if $x>0$ then $x\times [0,1]$ is a countable neighborhood of $x$ in $T$ and we can easily show that any continuous map $f$ from a unit segment $I=[0,1]$ (endowed with the standard topology) to $T$ such that $f(0)=x\times 0$ is constant. So maybe you have to consider on $T$ the order topology induced from $[0,1]^2$, maybe, even generated by a flipped order $a\times b< c\times d$ iff $b<d$ or $b=d$ and $a<c$.
• From the part "Also, something is wrong,..." --- this is exactly what I am trying to say. I knew that $T$ in the second case is not path connected. – topology_001 Sep 22 '17 at 6:33
• Actually I want to know why the argument for the first problem does not work in the next case : (i.e. why the argument when $X$ is countable does not work in the the case of when $X=[0,1]\times 0$. – topology_001 Sep 22 '17 at 6:34
• It is just: I am saying "consider any two point in $T$, since they are path connected with $p$, T is path connected". Why this argument do not work in the next case! – topology_001 Sep 22 '17 at 6:38
• .... So the conclusion is: If that argument do not work for $X=[0,1]\times 0$, how can I be sure that that argument will work for $X=$ rational points on $[0,1]\times 0$ .... I hope you understand!!! – topology_001 Sep 22 '17 at 6:41