0
$\begingroup$

The statement is the following: $X$ is normal $\Leftrightarrow$ for every two open sets $U,V\subseteq X$ with $U\cup V=X$, there exists closed sets $C\subseteq U$ and $D\subseteq V$ which also satisfies $C\cup D=X$.

So the in the notes the lecturer wrote that when proving this "(Careful: remember that we included Hausdorffness in the definition of normal, so you should verify this as well.)"

I think I have completed the proof, but should I show that X is Hausdorff aswell in the "$\Leftarrow$" or I am mistunderstanding?

  • 0
    Can you just verify that the definition of Hausdorff is satisfied? – 2017-01-11
  • 0
    I'm not sure how to tho – 2017-01-11
  • 1
    You're out of luck. The condition is just a rewriting of the standard formulation "every two disjoint closed sets are separated by open sets" of the $T_4$ separation axiom, and that doesn't imply Hausdorffness. It would if you are only considering $T_1$ spaces, but not in general. Every indiscrete space satisfies the condition, but if it contains more than one point is not Hausdorff. – 2017-01-11

1 Answers 1

1

If your lecturer includes Hausdorffness in the definition of normality, the result is false. Let $X=\Bbb N$. For each $n\in\Bbb N$ let $U_n=\{k\in\Bbb N:k

$$\tau=\{X\}\cup\{U_n:n\in\Bbb N\}$$

is a $T_0$ topology on $X$ that is not $T_1$, let alone Hausdorff. But if $V,W\in\tau$, and $V\cup W=X$, then one of $V$ and $W$ is $X$. Say $W=X$. Then $\varnothing\subseteq V$ and $X\subseteq W$ are closed sets whose union is $X$, so the righthand condition in the stated result is true. And indeed $X$ is normal in the preferred sense of the word, which does not include Hausdorffness: $X$ does not have two disjoint non-empty closed subsets, so $X$ is vacuously normal. (What your lecturer calls normal is better called $T_4$.)

  • 0
    Normal space definition: X is said to be normal if for each pair A,B of disjoint closed sets of X, there exists disjoint open sets containing A and B respectively. – 2017-01-11
  • 0
    @seht111: That is the preferred definition, and it is equivalent to the other condition in your question. However, your lecturer is defining it to be that **plus Hausdorffness**, or what I call $T_4$, which is *not* equivalent to the other condition. – 2017-01-11
  • 0
    I don't know if it means anything, but the definition also assumes that one-point sets are closed. – 2017-01-11
  • 0
    @seht111: I will use the preferred terminology, which differs from your lecturer’s. For me $X$ is normal if disjoint closed sets in $X$ have disjoint open nbhds. $X$ is $T_4$ if $X$ is normal and $T_1$. Apparently your lecturer has defined *normal* to mean what I would call $T_4$, normal plus $T_1$. This is equivalent to (my) normal plus Hausdorff. The point remains that the result is false for the definition that your lecturer has chosen to adopt. I gave you a counterexample, and I see that Daniel Fischer has given you another in the comments. – 2017-01-11
  • 0
    Let p : X → Y be a closed, continuous, surjective map. Show that if X is normal, then Y is normal as well. You can prove this in any way you like, but one suggestion is the following: first prove that Y is normal if and only if, for every two open sets U, V ⊆ Y with U ∪ V = Y , there exist closed sets C ⊆ U and D ⊆ V which also satisfy C ∪ D = Y . (Careful: remember that we included Hausdorffness in the definition of normal, so you should verify this as well.) So proving this wouldn't be possible if the definition contains Hausdorffness? – 2017-01-11
  • 0
    @seht111: This result is true even if Hausdorffness is included. Hausdorff implies $T_1$, so singletons in $X$ are closed. The map is closed, so singletons in $Y$ are closed. Now use continuity of $f$ and normality (without Hausdorffness) of $X$ to show that $Y$ is Hausdorff. – 2017-01-11
  • 0
    So what the lecturer ment, was that I had to show that Hausdorfness of X $\Rightarrow$ Hausdorffness of Y, and not what i originally wanted to show.? – 2017-01-11
  • 0
    @seht111: This is a completely different result from the one in your original question. The one in your question is false. This one is true, and yes, you have to show that $Y$ is Hausdorff in addition to showing that it’s what I call normal. – 2017-01-12
  • 0
    I'm just abit confused I guess. I know that the results are different, but is it possible to use the result in my original question, to prove this one? – 2017-01-12
  • 0
    @seht111: Of course not: the result in your original question is **false**. – 2017-01-12
  • 0
    @seht111: I should probably expand that last comment. It *is* true that normality in my sense (i.e., without necessarily requiring Hausdorffness) is equivalent to the other condition, and you will want that result if you intend to use the hint that you were given. – 2017-01-12
  • 0
    So would it be acceptable to use this result to show that Y is normal(in your sense) and then use that p is a closed map, that maps one-point sets in X,(which are closed), to closed one-point sets in Y, Y hence Hausdorff. – 2017-01-12
  • 0
    @seht111: Yes, and that’s probably what the hint was intended to suggest. – 2017-01-12