Hi

Do you know an exemple of connected space but path connected in no open subset (except perhaps the whole space) ?

Printable View

- Mar 3rd 2009, 12:50 PMclic-claca topological space
Hi

Do you know an exemple of connected space but path connected in no open subset (except perhaps the whole space) ? - Mar 7th 2009, 09:17 AMclic-clac
Could anyone tell me if that's a wrong exemple (but I think it works): in $\displaystyle \mathbb{R}^2,$

$\displaystyle (\mathbb{Q}\times\mathbb{R}\cup\mathbb{R}\times\{0 \})-\mathbb{Q}\times\{0\}$ - Mar 7th 2009, 11:20 AMOpalg
That example looks correct to me (but proving it could be quite messy—I'm glad I don't have to do that). (Wink)

(Also, a few extra parentheses would make it easier to read: $\displaystyle (\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\})$.) - Mar 7th 2009, 01:40 PMbkarpuz
$\displaystyle (\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\})=\mathbb{Q}^{c}\times(\math bb{R}\backslash\{0\})?

$

By the way, can you please give the definitions that you use?

*connected*and*path connected*? - Mar 7th 2009, 01:43 PMbkarpuz
- Mar 7th 2009, 11:41 PMclic-clac
Definitions I use is:

$\displaystyle X$ connected iff for all $\displaystyle U,V$ disjoint open subsets in $\displaystyle X,\ X=U\cup V\Rightarrow (U=\emptyset,\ V=X)$ or $\displaystyle (U=X,\ V=\emptyset)$

$\displaystyle X$ is path connected iff between any two points $\displaystyle x,y$ in $\displaystyle X$ there is always a path. (a continuous map $\displaystyle \sigma$ from $\displaystyle [0,1]$ to $\displaystyle X$ such that $\displaystyle \sigma(0)=x$ and $\displaystyle \sigma(1)=y$) - Mar 8th 2009, 12:58 AMbkarpuz
Why not only considering

$\displaystyle

\mathbb{Q}\times\mathbb{R},

$

which has the same nature with the first quadrant of

$\displaystyle

(\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\}).

$

It is clear that this set is not path connected.

Because the mapping $\displaystyle \sigma:[0,1]\to\mathbb{Q}\times\mathbb{R}$ will have discontinuity while matching $\displaystyle \sigma(0)=(q_{0},r_{0})$ and $\displaystyle \sigma(1)=(q_{1},r_{1})$ for some $\displaystyle P_{0}=(q_{0},r_{0})$ and $\displaystyle P_{1}=(q_{1},r_{1})$, for instance when $\displaystyle q_{0}\neq q_{1}$, and in any open subset you may find such points (may be sketching a graphic would help for this).

In this case, you will always have $\displaystyle \sigma(t)\in\mathbb{Q}^{c}\times\mathbb{R}$ for some $\displaystyle t\in[0,1]$.

About the connectivity, I am not sure for this set. - Mar 8th 2009, 03:20 AMclic-clac
You're right it's not path-connected, but unfortunately it's not connected: you can write this space as a disjoint union of non trivial open sets, for instance:

$\displaystyle \mathbb{Q}\times\mathbb{R}= ((\mathbb{Q}\cap ]-\infty ,\pi [)\times\mathbb{R}) \cup ((\mathbb{Q}\cap]\pi ,+\infty [)\times\mathbb{R})$

(this sets are open because they are intersections with open subsets in $\displaystyle \mathbb{R}^2$, and we're working with subspace topology)

But the space I was looking for may be more complicated: it has to be connected, but, instead of not path connected, not path-connected in any open subset (i.e. if you want never locally path-connected), and I'm not sure that ($\displaystyle X$ not path-connected)$\displaystyle \Rightarrow$($\displaystyle X$ not path-connected in any open subset)... - Mar 8th 2009, 03:39 AMbkarpuz
Okay, please see here Connected and Path Connected

I guess this will help you much, and indicates that your assertion is true.