2
$\begingroup$

Show an example of a path connected topological space that it's not locally connected.

Example (by William Eliot in this post Path connected space that is not locally connected):

In $\mathbb R^2$, draw lines from $(0,0)$ to $(1,q)$ for all rational $q$ in $[0,1]$
and from $(1,0)$ to $(0,-q)$ for all rational $q$ in $[0,1].$

This subspace is path connected and nowhere locally connected.

My questions,

  1. How to write in math notation the above example? I draw it and I think it has something to do with the lines

$$xt+(1-t)y, x\in[0,1]$$

  1. To prove that it's path connected, the function that I should consider is $f(t)=tx+(1-t)y,t\in[0,1].$ Am I correct?

  2. To prove that it's not locally connected should I use the definition? Can you give me a hint to prove this?

$\endgroup$
  • $\begingroup$ For 3, I would look at a neighborhood of (1,0). $\endgroup$ – Daniel Jan 5 '18 at 19:55
  • $\begingroup$ Drawing a picture might be more illuminating than writing everything in terms of functions too. $\endgroup$ – Daniel Jan 5 '18 at 19:56
  • 1
    $\begingroup$ That example is nowhere locally connected. An example that is path-connected but not everywhere locally connected is $A\cup B \cup C\cup D$ where $A=\{0\}\times [-1,2]$ and $B=\{(x,\sin 1/x): 0<x\leq \pi\}$ and $C=\{\pi\}\times [0,2]\}$ and $D= [0,2]\times \{2\}. $ $\endgroup$ – DanielWainfleet Jan 6 '18 at 16:59
  • $\begingroup$ @DanielWainfleet But in any case, the example work. Doesn't it? $\endgroup$ – Bellatrix Jan 6 '18 at 18:59
  • $\begingroup$ Yes, it does, and it might have a name but I can't recall. $\endgroup$ – DanielWainfleet Jan 6 '18 at 20:57
3
$\begingroup$

1.) This is already in math notation (words don't make it unmathematical). If you want you can write it as

$$ M:=\bigcup_{q\in \mathbb{Q}\cap [0,1]} \big(\{ (t, qt)\in \mathbb{R}^2 \ : \ t \in [0,1] \} \cup \{ (t, -q + qt) \in \mathbb{R}^2 \ : \ t\in [0,1] \} \big).$$

2.) You can connect the points $(0,0)$ and $(1,0)$ via the path $t\mapsto (t,0)$. The points in the upper halfplane are connected to $(0,0)$ via the path $t\mapsto (t, qt)$ and the points in the lower halfplane are connected to $(1,0)$ via the path $(t, -q + qt)$.

3.) Recall the definition of locally connected space. A space is called locally connected, if for every element $x$ in this space and every open neighborhood $U$ there exists an open connected set $V$ such that $x\in V \subseteq U$.

Our space is not locally connected. For this pick $x=(1,1)$ intersect $M$ with with the upper halfplane. Let now $x\in V\subseteq M \cap(\mathbb{R}_{>0} \times \mathbb{R})$ be an open connected neighborhood of $(1,1)$. For $r>0$ denote by $B_r(x)$ the ball of radius $r$ centered at $x$. As $V$ is open, there exists $1>r>0$ such that $B_r(x) \cap M \subseteq V$.

Now we prove that $V$ is not connected. Choose $\alpha \in (1-r,1]$ irrational. Then $ V $ can be written as the disjoint union of

$$ V_1 := V\cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y > \alpha x\} $$

and

$$ V_2=V\cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y < \alpha x\}. $$

This follows from the following observation:

$$ M \cap \{ (x,y) \in \mathbb{R} \times \mathbb{R} \ : \ y = \alpha x\} = \{ (0,0) \} $$

(check this). Thus, we have because $V\subseteq \mathbb{R}_{>0} \times \mathbb{R}$

$$ V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y = \alpha x\} = \emptyset $$

and hence

$$ V = V \cap (\mathbb{R}_{>0} \times \mathbb{R}) = (V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y > \alpha x\}) \cup (V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y < \alpha x\}) \cup (V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y = \alpha x\}) = (V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y > \alpha x\}) \cup (V \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y < \alpha x\}) \cup \emptyset = V_1 \cup V_2.$$

As $\{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y > \alpha x\} \cap \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y < \alpha x\} = \emptyset$, we have $V_1 \cap V_2 = \emptyset$. I.e. $V$ is the disjoint union of $V_1$ and $V_2$.

Note that $V_1,V_2$ are open as both $\{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y > \alpha x\}$ and $ \{ (x,y) \in \mathbb{R}_{>0} \times \mathbb{R} \ : \ y < \alpha x\} $ are open subsets of $\mathbb{R}\times \mathbb{R}$. As $B_r(x)\cap M \subseteq V$ we have

$$ x \in V_1, \qquad \left(\frac{1-r+\alpha}{2} ,1\right) \in V_2 $$ Thus, $V$ can be written as the disjoint union of two nonempty open sets and is therefore not connected.

Note that you can play the same game all the other points and thus, as pointed out by DanielWainfleet in the comments, our space is nowhere locally connected.

$\endgroup$
  • $\begingroup$ For 3.) It's because $\forall x,z\in\mathbb Q,\exists y\in\mathbb I,$ right? $\endgroup$ – Bellatrix Jan 6 '18 at 19:15
  • 1
    $\begingroup$ @Ella I'm sorry, I don't understand your question. Which part of 3.) is not clear? What do you mean by $\mathbb{I}$? $\endgroup$ – Severin Schraven Jan 6 '18 at 19:29
  • 1
    $\begingroup$ The point is that $M\cap \{(x, \alpha x ) \ : \ x \in \mathbb{R}\} = \{(0,0)\}$. But $(0,0)\notin M \cap (\mathbb{R}_{>0}\times \mathbb{R})$. Thus, the union of the two sets I wrote down really give $M \cap (\mathbb{R}_{>0}\times \mathbb{R})$. $\endgroup$ – Severin Schraven Jan 6 '18 at 19:43
  • 1
    $\begingroup$ @lomber Thank you very much. $\endgroup$ – Severin Schraven Apr 12 '18 at 20:02
  • 1
    $\begingroup$ @stupid Of course you are right. I should have written $(x,y)\in \mathbb{R}\times \mathbb{R}$. At the moment I do not have access to a proper computer and therefore I prefer not to correct it right now. I will never find the right spot on my tiny mobile screen. $\endgroup$ – Severin Schraven Sep 9 '18 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.