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I found this in a handwritten note:

Defination : A topological space $X$ is $T_1$ if $\forall x \neq y \in X$ there exist a neighborhood of $y$ such that s.t. $x \not\in V$.

I was almost certain that this was incorrect, because it looked like the definition for $T_0$ space.

However, subsequently this definition was used to prove that all singletons are closed, for example:

$(\Leftarrow)$ Suppose $\{x\}$ is closed, then $\{x\}^c$ is open and $y \in \{x\}^c$ satisfies the definition of a $T_1$ space.

Is this some alternative definition of $T_1$ space?

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The definition is correct, albeit a little sloppy. Here’s a slightly more careful version of it:

$X$ is $T_1$ if for each $x\in X$ and each $y\in X\setminus\{x\}$ there is a nbhd $V$ of $y$ such that $x\notin V$.

Let $p$ and $q$ be any two distinct points of $X$. If we set $x=p$ and $y=q$, we see that the definition ensures that $q$ has a nbhd that does not contain $p$. We can just as well set $x=q$ and $y=p$, however, and conclude that $q$ has a nbhd that does not contain $p$. Thus, the definition really does say that $X$ is $T_1$. The point is that since both quantifiers are universal, we can interchange the rôles of $x$ and $y$ in the definition.

The real problem with the quoted version is that it doesn’t make the quantifier on $x$ explicit. Because of this, it’s easy on first reading to get the impression that the definition is asymmetric in $x$ and $y$, like the definition of $T_0$ separation.

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  • $\begingroup$ That's interesting. Because what I have is that $(X, \tau)$ is $T_1$ if $\forall x \neq y \in X$, $\exists U, V \in \tau$ such that $x \in U, y \not\in U$ and $y \in V, x \not\in V$. Whereas in the def above $x$ is not enclused in another open set $\endgroup$ – Olórin Jun 14 '16 at 3:54
  • $\begingroup$ @MSEisadatingsite: You don't have to assert the existence of both $U$ and $V$ in that definition. If you assert only the existence of $U$, then you can get a set $V$ by just applying the definition again with $x$ and $y$ swapped. $\endgroup$ – Eric Wofsey Jun 14 '16 at 3:55
  • $\begingroup$ @MSEisadatingsite: That follows immediately from the rephrased version that I gave: it gives an open $V$ such that $y\in V$ and $x\notin V$, and after you interchange $x$ and $y$ it gives you an open $U$ such that $x\in U$ and $y\notin U$. $\endgroup$ – Brian M. Scott Jun 14 '16 at 3:55
  • $\begingroup$ But we cannot do the same for $T_2$ space right? We must show there exists two disjoint open sets $\endgroup$ – Olórin Jun 14 '16 at 3:56
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    $\begingroup$ @MSEisadatingsite: Correct: this style of definition of the $T_2$ property requires talking about both open sets at the same time, because they must be related in a certain way: specifically, they must be disjoint. Note, though, that there are ways to define $T_1$ and $T_2$ that look very different from these: you should try to prove that $X$ is $T_1$ if and only if $\{x\}$ is the intersection of all open nbhds $x$ for each $x\in X$, and $X$ is $T_2$ iff $\{x\}$ is the intersection of all closed nbhds of $x$ for each $x\in X$. $\endgroup$ – Brian M. Scott Jun 14 '16 at 3:59
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Yes, this definition is correct. It is well-known that a space is $T_1$ iff all singleton sets are closed. For the other direction of the proof, suppose $X$ is $T_1$, let $x\in X$, and let $U$ be the union of all open sets which do not contain $x$. Then $U$ is open since it is a union of open sets, and does not contain $x$. Since $X$ is $T_1$, for any $y\neq x$, there is an open $V$ such that $y\in V$ and $x\not\in V$. The set $V$ is then one of the sets whose union is $U$, so $V\subseteq U$. In particular, $y\in V$. Since $y\neq x$ was arbitrary, this means $U=X\setminus\{x\}$. Thus $X\setminus\{x\}$ is open, so $\{x\}$ is closed.

The difference between this definition and the definition of $T_0$ is that in the definition of $T_0$, all you know is that there exists an open set $V$ such that either $y\in V$ and $x\not\in V$ or $x\in V$ and $y\not\in V$. So for example, consider the space $S=\{0,1\}$ with topology $\{\emptyset,\{0\},\{0,1\}\}$. This space is $T_0$: the only pair of points with $x\neq y$ is $0$ and $1$, and $\{0\}$ is a set which contains one but not the other (for some ordering of the two points). However, this space is not $T_1$, since if you set $x=0$ and $y=1$, there is no open set which contains $y$ but not $x$.

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  • $\begingroup$ When there are two good answers which one should I accept? $\endgroup$ – Olórin Jun 14 '16 at 4:00
  • $\begingroup$ Whichever one you found more helpful! Personally, I think Brian Scott's answer is a bit better than mine, so I'd encourage you to accept it. $\endgroup$ – Eric Wofsey Jun 14 '16 at 4:01

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