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Thread: a topological space

  1. #1
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    a topological space

    Hi

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

    $\displaystyle (\mathbb{Q}\times\mathbb{R}\cup\mathbb{R}\times\{0 \})-\mathbb{Q}\times\{0\}$
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  3. #3
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    Quote Originally Posted by clic-clac View Post
    Could anyone tell me if that's a wrong exemple (but I think it works): in $\displaystyle \mathbb{R}^2,$

    $\displaystyle (\mathbb{Q}\times\mathbb{R}\cup\mathbb{R}\times\{0 \})-\mathbb{Q}\times\{0\}$
    That example looks correct to me (but proving it could be quite messyI'm glad I don't have to do that).

    (Also, a few extra parentheses would make it easier to read: $\displaystyle (\mathbb{Q}\times\mathbb{R}) \cup(\mathbb{R}\times\{0\})-(\mathbb{Q}\times\{0\})$.)
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  4. #4
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    $\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?
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  5. #5
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    Quote Originally Posted by bkarpuz View Post
    $\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?
    Okay I was wrong!
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  6. #6
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    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 $\displaystyle \sigma(1)=y$)
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  7. #7
    Senior Member bkarpuz's Avatar
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    Exclamation

    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.
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  8. #8
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    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)...
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  9. #9
    Senior Member bkarpuz's Avatar
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    Exclamation

    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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