0
$\begingroup$

Let $X$ be a topological space with the following property: for any open subset $A$ of $X$ there is a collection of clopen subsets $\{A_{\alpha} : \alpha\in S\}$ such that $\overline{A}=\overline{\bigcup A_{\alpha}}$ (where $\overline{A}$ denotes the closure of $A$). Can anyone find an example of a space which has the above property but is not zero-dimensional?

$\endgroup$
3
  • 2
    $\begingroup$ Motivation? This seems like a homework problem . . . $\endgroup$ Nov 23, 2012 at 21:02
  • $\begingroup$ You haven't specified which definition of dimension you're using. Have you looked in the standard sources for whichever definition you're using? I agree with Noah, the question seems poorly motivated. $\endgroup$ Nov 23, 2012 at 21:47
  • $\begingroup$ Clarification... One definition for "zero-dimensional" (small inductive dimension, I guess) says: there is a base for the topology consisting of clopen sets. We can rephrase that as: for any open set $A$ there is a colletion of clopen sets $\{A_{\alpha}: \alpha\in S\}$ such that ${A}={\bigcup A_{\alpha}}$. So clearlly such a space has the property in the question. But maybe adding the closure gets us additional spaces. $\endgroup$ Nov 23, 2012 at 21:56

3 Answers 3

4
$\begingroup$

Any space which contains a dense set of isolated points will have the property. It is easy to construct (e.g. as subspaces of $\mathbb{R}^n$) such spaces with arbitrarily large dimension. For a one dimensional explicit construction see B.M.Scott's answer to this question: https://math.stackexchange.com/questions/152390/cardinality-of-a-dense-open-set.

$\endgroup$
6
  • $\begingroup$ Dear Ramiro What you think about the following question? Is there a completely regular Hausdorff space $X$ which has my property(in main questio) but is not zero-dimensional? $\endgroup$
    – Ali
    Nov 24, 2012 at 12:55
  • $\begingroup$ Ali, subspaces of the plane are completely regular (in fact normal) Hausdorff spaces. $\endgroup$ Nov 24, 2012 at 13:08
  • $\begingroup$ Any space which contains a dense set of isolated points has my property but may be a zero-dimensional space so is not my answer (e.g, $\beta N$ ). $\endgroup$
    – Ali
    Nov 24, 2012 at 20:38
  • $\begingroup$ Note that Zero-dimesional is a $T_{1}$ space which has a base contains clopen subsets. $\endgroup$
    – Ali
    Nov 24, 2012 at 20:39
  • $\begingroup$ Ali, You must have misunderstood the answer that Ramiro links to. It gives an example of a subset in $\mathbb{R}^2$ that contains an interval and also a dense set of isolated points. (Take a line segment $I$ in $\mathbb{R}^2$, and add a sequence of points in $\mathbb{R}^2\setminus I$ that accumulates everywhere on $I$.) $\endgroup$ Nov 25, 2012 at 16:31
0
$\begingroup$

Yes, there are spaces where this property holds which are not zero-dimensional: consider the two-point space $\lbrace a, b\rbrace$ with topology $\lbrace \lbrace a\rbrace, \lbrace a, b\rbrace, \emptyset \rbrace$. Then this space clearly does not have a basis of clopen sets, so is not zero-dimensional, but the only closed sets are $\emptyset$ and $\lbrace a, b\rbrace$ which are both the closures of clopen sets.

Note that a strengthening of your property - where we require the clopen $A_\alpha$ to be subsets of $A$ - does not hold in this case, and I suspect that stronger version is equivalent to being zero-dimensional. [EDIT: as Ramiro points out, this is not the case; see comments]

Also, note that there are multiple definitions of "zero-dimensional," so you should specify which one you mean.

(I am still curious what the motivation for this problem is.)

$\endgroup$
3
  • $\begingroup$ Spaces with a dense set of isolated points also have the stronger property you propose. And there are plenty of non-zero-dimensional spaces with a dense set of isolated points. $\endgroup$ Nov 24, 2012 at 3:50
  • $\begingroup$ Dear Friends Thank you very much of your answers. A $T_{1}$-space $X$ which has a base consisting of clopen sets is called zero-dimensional space. $\endgroup$
    – Ali
    Nov 24, 2012 at 4:39
  • $\begingroup$ Given that you want your space to be $T_1$, my example doesn't apply; Ramiro's, on the other hand, appears to exactly answer your question. $\endgroup$ Nov 24, 2012 at 6:08
-1
$\begingroup$

It is Easy to see that $X$ has my property if and only if $\beta X$ has my property. So if we consider Dowkers example $Y$ then $Y$ is a zero-dimensional where $\beta Y$ is not zero-dimensional (see general topology book, Ryszard Engelking). On the other hand $\beta Y$ has my property.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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