17
$\begingroup$

Edit: According to the Gelfand duality between topological spaces and commutative $C^{*}$algebras, I add some new tags. So the question is that what is the structure of $ Ext (A,A)$ where $A$ is $C_{0}(\mathbb{R})$. One can think to stabilization of this question, that is $A=C_{0}(\mathbb{R})\otimes \mathcal{K} $ where $\mathcal{K}$ is the algebra of compact operators.

The following was the first version of my question:

Are there only a finite number of connected topological spaces $X$ (up to homeomorphism) with the property that $X$ has an open subset $U$ such that $U$ and $X-U$ are homeomorphic to $\mathbb{R}$? I know three examples as I wrote in the title of this question. (We delete the end critical points from each letter.) Among capital alphabet, there are no other topological type with the above property.

Is it true that any space $X$ with this property can be embedded in $\mathbb{R}^{2}$?

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ It's not clear why one would expect there to be only finitely many...for example consider the space $X\subset \mathbb R^2$ formed as the union of the $y$-axis and the graph of the function $f(x)=\sin{(1/x)}$ for $x>0$. It seems that one could construct infinitely many examples by allowing the lines to accumulate on each other in complicated ways like this. Likewise there doesn't seem to be a reason to expect them to all embed in $\mathbb R^2$. $\endgroup$
    – Mike Jury
    Commented Dec 19, 2014 at 14:43
  • 8
    $\begingroup$ +1 for misleading yet amusing question title $\endgroup$
    – Zubin Mukerjee
    Commented Dec 19, 2014 at 14:47
  • 3
    $\begingroup$ @ViditNanda: $U$ is assumed to be homeomorphic to $\mathbb R$, so it is not "arbitrary". Or didn't I understand your comment properly? $\endgroup$
    – Loïc Teyssier
    Commented Dec 19, 2014 at 15:39
  • 1
    $\begingroup$ Indeed, to avoid examples like Mike's you probably want something a bit stronger like local connectedness or path-connectedness (as well as the Hausdorff property). $\endgroup$
    – Ian Morris
    Commented Dec 19, 2014 at 16:16
  • 2
    $\begingroup$ By the way, why did you choose the word "ART" instead of "TAR" or "RAT"? :) $\endgroup$
    – Joonas Ilmavirta
    Commented Dec 21, 2014 at 18:47

2 Answers 2

Reset to default
5
$\begingroup$

Yes there are infinitely many, by a version of Mike Jury's idea. In fact there are uncountably many embeddable in $\mathbb R^2$. Take the union of two real curves:

the open one $y= x^{-1}\sin(1/x)$

the closed one $x= f(y)$ for $f$ a function that is $0$ on some closed set $S$ and negative elsewhere.

The closure of the open curve is exactly $S$, so we can identify $S$ from the topology of the space. Clearly there are uncountably many closed subsets of $\mathbb R$ up to homeomorphism of $\mathbb R$ (e.g. encode a real number as a sequence of closed intervals, isolated points, and Cantor sets.) So there are uncountably many spaces.

Improve this answer
$\endgroup$
6
  • $\begingroup$ My deep thanks for your beautiful idea.So it seems that $Ext(A,A)$ is rich. $\endgroup$
    – Ali Taghavi
    Commented Dec 21, 2014 at 18:39
  • $\begingroup$ You beat me to it! I was just about to write this same answer after my last comment. $\endgroup$
    – Joonas Ilmavirta
    Commented Dec 21, 2014 at 18:45
  • $\begingroup$ @WillSawin Thanks again for your answer. Can one introduce a counterexample of this situation(infinite number of such $X$) such that each $X$ is locally compact?(According to the first paragraph of my question, Gelfand duality betwee. LC space and Commutative algebra) $\endgroup$
    – Ali Taghavi
    Commented Dec 22, 2014 at 4:40
  • $\begingroup$ @WillSawin According to Gelfand duality, I should reconsider my statement that $Ext(A,A)$ is rich, so it would be interesting to think about the structure of this group. $\endgroup$
    – Ali Taghavi
    Commented Dec 22, 2014 at 6:43
  • 1
    $\begingroup$ @WillSawin No it is not closed unles $S$ is the whole "y-axis". $\endgroup$
    – Ali Taghavi
    Commented Dec 23, 2014 at 12:32
16
$\begingroup$

There is a nonmetrizable space $X$ with that property so the embedding property fails. I don't know if one could build more elaborate examples along these lines to find infinitely many such spaces. The answer to your question may be different if you require $X$ to be metrizable or Hausdorff or otherwise nice.

Let $A$ and $B$ denote two copies of $\mathbb R$ with the usual topology and let $X$ be the disjoint union of $A$ and $B$. Define a topology on $X$ so that $\alpha\cup\beta$ with $\alpha\subset A$ and $\beta\subset B$ is open iff

  • $\alpha$ and $\beta$ are open (in the usual real topology) and
  • $\beta=B$ or $\alpha=\emptyset$.

This is indeed a topology. The union of any two disjoint nonempty open sets does not meet $A$, so $X$ is connected. The topology is not Hausdorff and thus not metrizable; two distinct points in $A$ cannot have disjoint neighborhoods.

Now $B\subset X$ is open and its complement is $A$, and both of these subspaces are homeomorphic to the real line.

Improve this answer
$\endgroup$
4
  • $\begingroup$ A question on haussdorf case: Assume that $X$ is haussdorf and satisfies the condition of my question. Is $X$ necessarilly locally compact? $\endgroup$
    – Ali Taghavi
    Commented Dec 21, 2014 at 11:10
  • $\begingroup$ @AliTaghavi, you're most welcome. I did enjoy constructing that example space. My guess is that a Hausdorff $X$ with your property is locally compact but I haven't given it a proper thought. $\endgroup$
    – Joonas Ilmavirta
    Commented Dec 21, 2014 at 17:23
  • $\begingroup$ I think the examples of Will Sawin are not locally compact. $\endgroup$
    – Ali Taghavi
    Commented Dec 22, 2014 at 4:42
  • $\begingroup$ @AliTaghavi, you are right. Local compactness seems to fail at the boundary of the accumulation set $S$. $\endgroup$
    – Joonas Ilmavirta
    Commented Dec 22, 2014 at 9:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .