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Need feedback. Especially on writing style.

QUESTION: Consider a topological space $(X, \mathcal{T} )$. Show that the following properties are equivalent.
(A) $X$ is Hausdorff.
(B) Let $x\in X$. For every $y\in X$ such that $x\ne y$, there exists $U(x)\in \mathcal{T}$ such that $y \notin \overline U(x)$.

(C) For $x \in X$, $\bigcap_{U \in \mathcal{U}} \bar U = \{x\}$

ANSWER: Suppose $(X, \mathcal{T})$ is a topological space. Let $\mathcal{U}=\{U \mid U \in \mathcal{T}, x \in U\}$ denote the family of all open neighbourhoods of $x$ in $X$. We show that $(A)$ and $(B)$ are equivalent.

Let $X$ be Hausdorff. Consider two arbitrary points $x,y$ in $X$, $x \neq y$. Since $X$ is Hausdorff and $x \neq y$, there exists an open neighbourhood $U \in \mathcal{U}$ of $x$ and an open neighbourhood $V \in \mathcal{T}$ of $y$, such that $U \cap V = \emptyset$. It follows that $y$ is not a point of closure of $U$ and thus, $y \notin \bar U$. Hence, we have shown that for every pair of distinct points $x,y \in X$, where $X$ is Hausdorff, there exists $U \in \mathcal{U}$ such that $y \notin U$.

Conversely, let, for every $x,y \in X$ such that $x \neq y$, there exist $U \in \mathcal{U}$ such that $y \notin \bar U$. This implies that $y$ is not a point of closure of $U$, and so there must exist an open neighbourhood of $y$, $V \in \mathcal{T}$ such that $U \cap V= \emptyset$. Thus, for every pair of distinct points $x,y \in X$, we have found a set of disdoint neighbourhoods, $U, V$ in $X$. Hence, $X$ is Hausdorff.

Now we show that (B) and (C) are equivalent. Let $\mathcal{U}$ be the family of all open neighbourhoods of $x$ in $X$, as already defined. Assume that for every $x,y \in X$ such that $x \neq y$, there exists $U \in \mathcal{U}$ such that $y \notin \bar U$. It follows from assumption that for any $y \neq x$, $y \notin \bigcap_{U \in \mathcal{U}} \bar U$. We are now required to show that $x \in \bar U$ for each $U \in \mathcal{U}$. Let $U,U' \in \mathcal{U}$ be arbitrary open neighbourhoods of $x$. Clearly, $x \in U'$ and $x \in U$, and so $x \in U \cap U'$. Consequently $U \cap U' \neq \emptyset$ and thus $x$ is a point of closure of each $U \in \mathcal{U}$. Therefore, $x \in \bigcap_{U \in \mathcal{U}} \bar U$. Hence, we have shown that $\bigcap_{U \in \mathcal{U}} \bar U = \{x\}$. Conversely, let $\bigcap_{U \in \mathcal{U}} \bar U = \{x\}$. It follows that for every $y \neq x$, there is some $U$ in $\mathcal{U}$ such that $y \notin \bar U$ and so, every $y \neq x$ is not a point of closure of $U$ for some (or the other) $U \in \mathcal{U}$. Thus, for every $y, y \neq x$, there exists $U \in \mathcal{U}$ such that $y \notin \bar U$.

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In the converse direction, why does $y\notin U$ imply that $y$ is not a point of closure of $U$? –  Thomas E. Feb 21 '13 at 13:34
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I think that you’ve misstated (B): it should almost certainly be the for each $y\in X\setminus\{x\}$ there is an open nbhd $U$ of $x$ such that $y\notin\operatorname{cl}U$. You’re missing the closure, though it appears sporadically in your ANSWER. The ANSWER contains a major error: $y\notin U$ does not imply that $y$ is not a point of the closure of $U$. –  Brian M. Scott Feb 21 '13 at 13:35
    
Sorry, I've just picked up LaTeX and I end up making lots of errors. –  Fatsho Feb 21 '13 at 13:44

2 Answers 2

$\newcommand{\cl}{\operatorname{cl}}$I’ll do what I did before: comment on the proof a bit at a time.

ANSWER: Suppose $(X, \mathcal{T})$ is a topological space. Let $\mathcal{U}=\{U \mid U \in \mathcal{T}, x \in U\}$ denote the family of all open neighbourhoods of $x$ in $X$. We show that $(A)$ and $(B)$ are equivalent.

This is fine, though I would write $\mathcal{U}=\{U\in\mathcal{T}:x\in U\}$ instead of putting both conditions on the righthand side of the such that divider.

Let $X$ be Hausdorff. Consider two arbitrary points $x,y$ in $X$, $x \neq y$. Since $X$ is Hausdorff and $x \neq y$, there exists an open neighbourhood $U \in \mathcal{U}$ of $x$ and an open neighbourhood $V \in \mathcal{T}$ of $y$, such that $U \cap V = \emptyset$.

There’s nothing really wrong with this, but it’s redundant to say both that $U$ is an open nbhd of $x$ and that it’s in $\mathcal{U}$: after all, if it’s in $\mathcal{U}$, by definition it’s an open nbhd of $x$. Similarly, it’s redundant to say that $V\in\mathcal{T}$ when you’re already said that $V$ is open. You could reduce the whole thing to this:

Let $X$ be Hausdorff. Consider arbitrary points $x,y\in X$ such that $x\ne y$. Since $X$ is Hausdorff and $x\ne y$, there are a $U\in\mathcal{U}$ and an open nbhd $V$ of $y$ such that $U\cap V=\varnothing$.

The next bit is fine until the typo at the very end:

It follows that $y$ is not a point of closure of $U$ and thus, $y \notin \bar U$. Hence, we have shown that for every pair of distinct points $x,y \in X$, where $X$ is Hausdorff, there exists $U \in \mathcal{U}$ such that $y \notin U$.

You want to conclude that $y\notin\overline U$.

Conversely, let, for every $x,y \in X$ such that $x \neq y$, there exist $U \in \mathcal{U}$ such that $y \notin \bar U$.

The English is a bit mangled here. You want something like this:

Conversely, suppose that for every $x,y\in X$ such that $x\ne y$ there is a $U\in\mathcal{U}$ such that $y\notin\overline U$.

The rest is basically fine, though there’s a little redundancy that can be removed. You have:

This implies that $y$ is not a point of closure of $U$, and so there must exist an open neighbourhood of $y$, $V \in \mathcal{T}$ such that $U \cap V= \emptyset$.

You could say simply

... an open nbhd $V$ of $y$ such that $U\cap V=\varnothing$.


By the way, (B) is easily seen to be equivalent to the assertion that for each $x\in X$,

$$\{x\}=\bigcap\{\cl U:x\in U\in\mathcal{T}\}\;.$$

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That's actually statement (C) of this question. :) –  Fatsho Feb 21 '13 at 14:48
    
@ Brian: I've just posted that proof as an edit. I don't suppose you could give me feedback on that as well? –  Fatsho Feb 21 '13 at 15:29
    
@Fatsho: It’s correct, but you’re working way too hard in the first half to show that $x$ is in the intersection. By definition $x\in U\subseteq\overline U$ for each $U\in\mathcal{U}$, so clearly $x\in\bigcap\mathcal{U}\subseteq\bigcap_{U\in\mathcal{U}}\overline U$. –  Brian M. Scott Feb 21 '13 at 15:35

The proof seems fine in general, though you can slightly simplify the converse direction by noting that if $y \notin \overline{U}$, then $V = X \setminus \overline{U}$ is an open neighbourhood of $y$ disjoint from $U$.

(Also, you are missing the closure-line above the "$U$" in the last line of the second paragraph of the proof.)

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Could you clarify the notation $V= X$ \ $\bar U$ for me? –  Fatsho Feb 21 '13 at 14:49
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@Fatsho: $V$ is the complement in $X$ of the closure of $U$. –  Brian M. Scott Feb 21 '13 at 14:52

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