Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think I have solved a problem in Topology by Munkres, but there is a small detail that is bugging me. The problem is stated in this question's title. I will write down the proof and will highlight what is troubling me.

We prove by contradiction: Assume $X$ is not Hausdorff. Then there exist points $x,y$ where $x$ is different from $y$ such that no neighbourhoods $U$, $V$ about $x$ and $y$ respectively have trivial intersection. Now consider the point $(x,y)$ that is in the complement of $\Delta$. Now let $U \times V$ be any basis element that contains $(x,y)$ (such an element exists by definition of the product topology being generated by the basis $\mathcal{B}$ consisting of elements of the form $W \times Z$, where $W$ is open in $X$ and $Z$ is open in $Y$). Consider $(U \times V ) \cap \Delta$, which I claim to be $(U \cap X) \times (V \cap X)$.

By our choice of $x$ and $y$ there is $z \in U \cap V$, implying that the intersection $(U \times V ) \cap \Delta$ is not trivial.

Since $U \times V$ was any basis element containing $(x,y)$, this means that $(x,y) \in \overline{\Delta}$, which means that there exists a limit point of $\Delta$ that is not in it, contradicting $\Delta$ being closed.

The problem comes is in the way I have decomposed $\Delta$; the way I have put it seems I am saying that $\Delta$ is equal to $X \times X$, which is not the case. How can I get round this?

Thanks.

Edit: Martin Sleziak has pointed out some mistakes, $(U \times V ) \cap \Delta$ should be $\{ (x,x) : x \in U \cap V\}$ and not as claimed.

share|improve this question
4  
See also this thread for a nice and useful generalization. –  t.b. Jan 2 '12 at 14:51
add comment

4 Answers

up vote 4 down vote accepted

Your claim that $(U\times V)\cap \Delta=(U\cap X)\times(V\cap X)$ is incorrect. Since $U,V\subseteq X$, this is the same as claiming $(U\times V)\cap \Delta=U\times V$.

If you use $(U\times V)\cap \Delta = \{(x,x); x\in U\cap V\}$ instead, the rest of your proof should work fine, but there are still several minor details.


Minor nitpicks:

  • Wouldn't a direct proof (instead of using contradiction) more clear? I don't think you would have to change the proof much to do this. But perhaps this is a matter of taste.

  • If you're using proof by contradiction, you cannot choose $x$, $y$ arbitrarily, but you should choose $x,y\in X$, $x\ne y$, which witness that this space is not Hausdorff. (I.e., for any neighborhoods $U\ni x$, $V\ni y$ the intersection $U\cap V$ is non-empty.)

One more nitpick considering formatting:

  • It's not good to write two formulas after each other. If you write "$x,y$ $x \neq y$ such that..." this is quite difficult to read. You should separate such things at least with a little, e.g. "$x$, $y$; $x \neq y$ such that..."; in my opinion it is much better to separate them with text, e.g. "$x$, $y$ such that $x \neq y$ and..."
share|improve this answer
    
That's what I wanted to say, you have put it in a better way. I will edit my post now. Thanks. –  fpqc Jan 2 '12 at 14:37
    
Bad slip up, I have corrected the thing on the quantifiers. –  fpqc Jan 2 '12 at 14:43
    
@Benjamin I've added one more comment on your formatting (LaTeX). –  Martin Sleziak Jan 2 '12 at 14:48
    
I have edited that too. –  fpqc Jan 2 '12 at 14:51
add comment

A straightforward proof is a bit simpler:

If the diagonal is closed, then for $x\ne y$ in $X$, the point $(x,y)$ is not on the diagonal. So, there is an nhood of $(x,y)$ in $X\times X$ containing $(x,y)$ disjoint from the diagonal. But, then, this nhood is of the form $U\times V$ where $U$ and $V$ are open disjoint sets containing $x$ and $y$ respectively.


Actually, contraposition would be best as you started in my opinion. If $X$ is not Hausdorf then there are $x$ and $y$ with $x\ne y$ such that for every open set $U$ containing $x$ and for every open set $V$ containing $y$ we have $U\cap V\ne\emptyset$ . Then, as you show, $(x,y)$ is in the closure of the diagonal. So, as $(x,y)$ is not on the diagonal but is in its closure, the diagonal is not closed.


The answer to your actual question is nicely explained in the other posts here. But I wanted to emphasise that an nhood $U\times V$ in $X\times X$ (which is as you describe) intersects the diagonal if and only if $U\cap V\ne\emptyset$. This follows from the relevant definitions is the key to the whole problem...

share|improve this answer
add comment

You made the statement "Since $X$ is not Hausdorff, for any neighbourhoods $U$ and $V$ about $x,y$ respectively the intersection of $U$ and $V$ is not trivial." This actually has to only be true for a single $(x,y)$, not all $x$ and $y$.

It's probably best to not use proof by contradiction... If $\Delta$ is closed, then for any $(x,y) \notin \Delta$, there are open $U$ and $V$ such that $(x,y) \in U \times V$ and $U \times V$ is disjoint from $\Delta$. If you translate this correctly, you're done.

share|improve this answer
    
When dealing with metric spaces, that's the way I would call an open set (a subset $E$ of a metric space $X$ is open iff every point is an interior point). This is in fact easier because $\Delta^{c}$ open iff $\Delta^{c} = \text{int} \Delta^{c}$ iff any $x \in \Delta^{c}$ is in some open neighbourhood entirely contained in $\Delta^{c}$. –  fpqc Jan 2 '12 at 14:49
    
It's about the definition of Hausdorff: "For every $x$ and $y$ in the space so and so holds". So the negation would be "There exists $x$ and $y$ such that so and so does not hold". In other words, it wouldn't have to be true for every pair $x$ and $y$, just one pair. –  Zarrax Jan 2 '12 at 14:52
    
I realised this mistake immediately once it was pointed out. –  fpqc Jan 2 '12 at 14:53
add comment

The definition of Hausdorff is that for all distinct pairs $x,y \in X$ there exist disjoint open $U \ni x$ and $V \ni y$. Hence the negative is that there exist $x,y \in X$ each of whose open neighbourhoods $U \ni x$, $V \ni y$ have nonempty intersection. You seem to have said that this is the case for all $x,y \in X$, which isn't true.

Also your claim $(U \times V) \cap \Delta = (U \cap X) \times (V \cap X)$ seems to assume $\Delta = X \times X$, which could be the cause of your problems.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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