3
$\begingroup$

I am trying to learn topology from these lecture notes (or look here for a longer version).

I was having some trouble with the difference between the definition of a $T_0$-space and a $T_1$-space. Eventually, with the help of this question I figured out the difference between the two. However I wasn't able to figure out the difference between the definitions given in the lecture notes. In the notes, a $T_0$-space is defined as a topological space $(X,\tau)$ such that (and I quote):

Given $x,y\in X, x\neq y$, there is a $U\in\tau$ containing precisely one of the two points. (I.e. all points are distinguished by $\tau$.)

On the other hand, a $T_1$-space is defined as:

For any $x,y\in X$ with $x\neq y$ there is an open set $U$ such that $x\in U$, $y\notin U$.

Now, I would interpret these definition as follows: for a $T_0$-space I would write $$\textbf{Given } x,y\in X: x\neq y\exists U\in\tau: x\in U,y\notin U,$$ and for a $T_1$-space: $$\textbf{For any } x,y\in X: x\neq y\exists U\in\tau: x\in U,y\notin U.$$ At first glance, looking only at these epxressions, I would say these definitions are identical, and replace both 'Given' and 'For any' with a '$\forall$' symbol. I know the crux of the definition of the $T_1$-space is that you can switch $x$ and $y$, which means you can find, at least, two open sets; one containing only $x$ and one containing only $y$. This is not the case for a $T_0$-space, where we can only find (at least) one open set; containing only $x$ and not $y$. But, in the definition given above I see no reason why we can't switch $x$ and $y$, just as we did in the $T_1$ case. (Again, see this question.)

Am I misunderstanding the definitions? Is this (symbolic) interpretation of the definitions correct? And if so, what is the difference exactly between the words 'given' and 'for any' which make the definitions different?

Thanks so much in advance!

$\endgroup$
  • 1
    $\begingroup$ containing precisely one of the two points it can happen that $x\notin U,y\in U,$ while in a $T_1$ space both situations must occur. $\endgroup$ – awllower Jul 29 '16 at 13:12
  • $\begingroup$ But in the $T_1$ definition it is also the case that $U$ msut contain 'precisely one of the two points', right? $\endgroup$ – Nesta Jul 29 '16 at 13:20
  • 1
    $\begingroup$ No, it specifically says $x\in U,y\notin U.$ If we take $y,x$ instead, then the condition $y\in U,x\notin U$ also holds. Well, just see Asaf's answer. :) $\endgroup$ – awllower Jul 29 '16 at 13:40
4
$\begingroup$

The difference is not in the "given" and "for any". The difference is that in $T_0$ we only require one of the two points to be in $U$, and we do not specify which one it can be; and in $T_1$ we require $x$ to be in $U$ (and by symmetry it means there is some open $V$ such that $y\in V$ and $x\notin V$).

Your interpretation is wrong, if so, of $T_0$. Because $T_0$ means $$(x\in U\land y\notin U)\lor(x\notin U\land y\in U).$$


Consider the topology on $\Bbb R$ defined by $$\{\varnothing\}\cup\{U\cup\{0\}\mid U\text{ open in the standard topology}\}.$$ This is a $T_0$ topology (why?), but it is not $T_1$ because while we can find an open set $U$ such that $0\in U$ and $1\notin U$, we cannot find an open set which gives us the other way around.

$\endgroup$
  • $\begingroup$ The line "...and we do not specify which one it can be..." made it click. In the definition I quoted it says "containing precisely one of the two points" (which does not specify which of the two points it should be), but in my interpretation I did specify which of the two points by stating that it should be $x$ which is contained in $U$. This is of course only one possibility. Thank you! $\endgroup$ – Nesta Jul 29 '16 at 13:43
3
$\begingroup$

Your interpretation of the definition of $T_0$-space is wrong: it does not say that

Given $x,y \in X$ if $x \ne y$ then there exists a $U \in \tau$ such that $x \in U$ and $y \not \in U$

instead it says that

Given $x,y \in X$ if $x \ne y$ then there is a $U \in \tau$ such that either $x \in U$ and $y \not \in U$ or $y \in U$ and $x \not \in U$.

The difference with $T_1$-spaces is that in a $T_1$-space

Given $x, y \in X$ with $x \ne y$ there are two open sets $U,V \in \tau$ such that $x \in U$ and $y \not \in U$ and $y \in V$ and $x \not \in V$.

In the first case we require the existence of an open set for at least one point, in the second we require the existence of an open set for each point.

Hope this helps.

$\endgroup$
2
$\begingroup$

"Given", "for any" and "for all" are the same. Your misconception is not in the quantifier, but in the interpretation of "$U$ contains exactly one of the two".

$T_0$ should be: $$\forall x,y\in X, (x\ne y\longrightarrow(\exists U\in\tau,\ ((x\in U\wedge y\notin U)\vee (y\in U\wedge x\notin U))))$$

While $T_1$ is indeed as you said.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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