a topological space

**clic-clac** · March 3rd 2009, 12:50 PM

Hi

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

**clic-clac** · March 7th 2009, 09:17 AM

Could anyone tell me if that's a wrong exemple (but I think it works): in $\mathbb{R}^2,$

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

**Opalg** · March 7th 2009, 11:20 AM

Originally Posted by clic-clac

Could anyone tell me if that's a wrong exemple (but I think it works): in $\mathbb{R}^2,$

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

That example looks correct to me (but proving it could be quite messy—I'm glad I don't have to do that).

(Also, a few extra parentheses would make it easier to read: $(\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\})$ .)

**bkarpuz** · March 7th 2009, 01:40 PM

$(\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?

**bkarpuz** · March 7th 2009, 01:43 PM

Originally Posted by bkarpuz

$(\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?

Okay I was wrong!

**clic-clac** · March 7th 2009, 11:41 PM

Definitions I use is:
$X$ connected iff for all $U,V$ disjoint open subsets in $X,\ X=U\cup V\Rightarrow (U=\emptyset,\ V=X)$ or $(U=X,\ V=\emptyset)$

$X$ is path connected iff between any two points $x,y$ in $X$ there is always a path. (a continuous map $\sigma$ from $[0,1]$ to $X$ such that $\sigma(0)=x$ and $\sigma(1)=y$ )

**bkarpuz** · March 8th 2009, 12:58 AM

Why not only considering
$ \mathbb{Q}\times\mathbb{R}, $
which has the same nature with the first quadrant of
$ (\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 $\sigma:[0,1]\to\mathbb{Q}\times\mathbb{R}$ will have discontinuity while matching $\sigma(0)=(q_{0},r_{0})$ and $\sigma(1)=(q_{1},r_{1})$ for some $P_{0}=(q_{0},r_{0})$ and $P_{1}=(q_{1},r_{1})$ , for instance when $q_{0}\neq q_{1}$ , 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 $\sigma(t)\in\mathbb{Q}^{c}\times\mathbb{R}$ for some $t\in[0,1]$ .
About the connectivity, I am not sure for this set.

**clic-clac** · March 8th 2009, 03:20 AM

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:

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

(this sets are open because they are intersections with open subsets in $\mathbb{R}^2$ , 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 ( $X$ not path-connected) $\Rightarrow$ ( $X$ not path-connected in any open subset)...

**bkarpuz** · March 8th 2009, 03:39 AM

Okay, please see here Connected and Path Connected
I guess this will help you much, and indicates that your assertion is true.

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