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Let $X$ be the real with the topology $T = \{ U \subset\mathbb{R}\:|\:0\notin U\text{ or }\mathbb{R}\setminus U\text{ finite}\} $. What is the largest value for $n \in \{0,1,2,3,3\frac{1}{2},4\}$ such that $X$ satisfies the separation axiom $T_n$?

First of all, the open sets in this topology are all cofinite sets and open intervals $I \subset \mathbb{R} \setminus \{0\}$. I would argue that the topology is hausdorff, by using only the open intervals (without using the cofinite sets) to find open neighborhoods $U$ and $V$ for some $a,b \in X$. But I'm not sure if I could go any higher into $T_3$.

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You need a bit more care for the Hausdorff property, not all open intervals are in your topology. –  Michael Greinecker Jan 14 '13 at 14:32
    
That works of $a,b\in X\setminus \{0\}$ but what if $a=0$? –  Thomas Andrews Jan 14 '13 at 14:34
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@MichaelGreinecker. Indeed, all open intervals not containing $0$ are excluded. But that does not change that for every point $p$ in $\mathbb{R} \setminus \{0\}$ there is an open inteval containing $p$. If this is right, isn't this also the property which quarantees Hausdorff? –  omar Jan 14 '13 at 14:41
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Suppose $a=0$, then the only possible open sets containing $a$ are the cofinite sets. Let $U$ be some open set such that $a \in U$ and $X \setminus U$ finite. Is there then still any point left which has a disjunct open neighborhood $V$? Because Hausdorff requires $U$ and $V$ disjunct it should be that $V \subset X \setminus U$; but that's not possible..? –  omar Jan 14 '13 at 15:04
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@omar Wlog let $x\neq 0$. Then $\{x\}$ and $\mathbb{R}\backslash\{x\}$ separate $x$ and $y$. –  Michael Greinecker Jan 14 '13 at 15:12

2 Answers 2

up vote 3 down vote accepted

No, the open sets are all subsets of $\Bbb R\setminus\{0\}$ together with all cofinite sets. You’re right that $X$ is Hausdorff, but since you described the topology incorrectly, you need to think a little more about how to show this.

Note that $X$ is compact. If $\mathscr{U}$ is an open cover of $X$, there is some $U_0\in\mathscr{U}$ such that $0\in U_0$. Now $X\setminus U_0$ is finite, so only finitely other elements of $\mathscr{U}$ are needed to complete a subcover.

HINT: What do you know about the separation properties of compact Hausdorff spaces?

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Well, compact Hausdorff spaces have some properties yes.. But am I allowed to utilize such properties when trying to prove some space Hausdorff? I would think that one should just try covering all possibilities for disjunct open neighborhoods for any two points? –  omar Jan 14 '13 at 14:58
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@omar: Obviously you can’t use those properties to prove that $X$ is $T_2$, and I wasn’t suggesting that you should. After you prove that $X$ is $T_2$, which does indeed require looking at two cases, then you use what you know about compact $T_2$-spaces to say something about the higher separation properties. –  Brian M. Scott Jan 14 '13 at 15:07
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Corollary: If $A$ and $B$ are disjunct closed subsets of a compact Hausdorff space $X$, then there exist disjunct open sets $U$ and $V$ in $X$ such that $A\subset U$ and $B\subset V$. –  omar Jan 15 '13 at 11:27
    
@omar: Yes: it’s $T_4$. –  Brian M. Scott Jan 15 '13 at 18:04

The possible values of $n$ are 0,1.

Proof of $X$ satisfies $T_1$ axiom: Let $x,y\in X$, $x\neq y$. Then $$U=X\setminus \{y\}, \quad V=X\setminus \{x\}$$ satisfies $x\in U\not\ni y$, $y\notin V \ni y$. And $U$ is neighborhood of $x$ and $V$ is neighborhood of $y$.

Proof of $X$ is not $T_2$ : Let $x,y\in X$, $x\neq y$. If $U$ is neighborhood of $X$ such that does not contain element $y$. If $y$ has neighborhood $V$ disjoint form $U$, We get $V$ is uncountable and $V\subset X\setminus U$. But $V\subset X\setminus U$ is finite and we get contradiction.

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Wlog let $x\neq 0$. Then $\{x\}$ and $\mathbb{R}\backslash\{x\}$ separate $x$ and $y$. –  Michael Greinecker Jan 14 '13 at 14:40

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