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?
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2$\begingroup$ Motivation? This seems like a homework problem . . . $\endgroup$– Noah SchweberCommented Nov 23, 2012 at 21:02
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$\begingroup$ 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. $\endgroup$– Ryan BudneyCommented Nov 23, 2012 at 21:47
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$\begingroup$ 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. $\endgroup$– Gerald EdgarCommented Nov 23, 2012 at 21:56
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3 Answers
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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.
answered Nov 23, 2012 at 23:24
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$\begingroup$ 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? $\endgroup$– AliCommented Nov 24, 2012 at 12:55
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$\begingroup$ Ali, subspaces of the plane are completely regular (in fact normal) Hausdorff spaces. $\endgroup$ Commented Nov 24, 2012 at 13:08
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$\begingroup$ 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, $\beta N$ ). $\endgroup$– AliCommented Nov 24, 2012 at 20:38
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$\begingroup$ Note that Zero-dimesional is a $T_{1}$ space which has a base contains clopen subsets. $\endgroup$– AliCommented Nov 24, 2012 at 20:39
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$\begingroup$ 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 $I$.) $\endgroup$ Commented Nov 25, 2012 at 16:31
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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.)
answered Nov 24, 2012 at 1:06
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$\begingroup$ 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. $\endgroup$ Commented Nov 24, 2012 at 3:50
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$\begingroup$ 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. $\endgroup$– AliCommented Nov 24, 2012 at 4:39
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$\begingroup$ 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. $\endgroup$ Commented Nov 24, 2012 at 6:08
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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.