# Exercises about Hausdorff spaces

two problems from Dugundji's book page $156$. (I don't know why the system deletes the word hi in this sentence)

$1$. Let $X$ be a Hausdorff space. Show that:

a) $\bigcap \{F: x \in F , F \ \textrm{closed}\} = \{x\}$.

b) $\bigcap \{U : x \in U, U \ \textrm{open}\} = \{x\}$.

c) The above properties are not equivalent to being Hausdorff.

$2$. Prove every infinite Hausdorff space contains a countably infinite discrete subspace.

My work:

1. (a) So let $y \not \in \{x\}$ then $x \neq y$. Since $X$ is Hausdorff we can find disjoint open sets $U_{x}$, $V_{y}$. Define $F = X \setminus V_{y}$ then $x \in F$, $F$ is closed and $y \not \in F$, so $y \not \in \bigcap \{F: x \in F , F \ \textrm{closed}\}$.

(b) Same as above, just note that $y \not \in U_{x}$, hence not in the intersection.

(c) Not sure here, can we take $X$ any infinite set endowed with the cofinite topology?

Claim: $\bigcap \{F: x \in F , F \ \textrm{closed}\} = \{x\}$. Suppose $y \not \in \{x\}$ then $y \neq x$, that is: $y \not \in \{x\}$. Note that since $X$ is a $T_{1}$ space then $F=\{x\}$ is closed and contains $x$, so $y$ is not in the intersection.

Similarly for $(b)$. But $X$ is not Hausdorff since any two non-empty open sets intersect.

2) How do you prove this one?

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Note that $\{x\}$ means the set containing $x$ only. So 1(a) should start with "Suppose that $y \ne x$." Rest of 1(a) looks fine. For 1(c), almost all of first paragraph not needed. Complement of $\{x\}$ is cofinite, so open. Need somewhat different but equally easy argument for the second property. –  André Nicolas Jul 7 '11 at 5:48

## 3 Answers

Some style comments on 1(a): As user6312 says, you should just begin by supposing that $y \ne x$; starting with $y \notin \{x\}$ is an unnecessary complication. More important, what Hausdorffness gives you is not just disjoint open sets $U_x$ and $V_y$, but disjoint open sets $U_x$ and $V_y$ such that $x \in U_x$ and $y \in V_y$, and for clarity you should say so. (Minor point: if you use different letters for the sets, there's no reason to use subscripts: just call them $U$ and $V$. If you use the subscripts, you might as well use a single basic name, e.g., $U_x$ and $U_y$.)

1(c): Just a brief expansion of what user 6312 said: $\{x\}$ is finite, so by definition its complement is open, and $\{x\}$ is therefore one of the closed sets containing $x$, and you don't need to look further to see that $\{x\}$ is the intersection of the closed sets containing $x$. Alternatively, if you've already proved that (a) is true in any $T_1$ space, you merely observe that if $x \ne y$, then $X \setminus \{y\}$ is an open set containing $x$ but not $y$, so $X$ is $T_1$. (But in that case you could have done (a) simply by noting that every Hausdorff space is $T_1$.) For the second part of (c) you do need the argument that $x$ is the intersection of the open sets $X \setminus \{y\}$ for $y \ne x$.

(2) If $x \in X$ has a finite open nbhd, then $\{x\}$ is open (i.e., $x$ is an isolated point). (Why?) Let $I$ be the set of isolated points of $X$; clearly $I$ is discrete, so if $I$ is infinite, just pick any countably infinite subset of it. That's the easy case; you have to work harder when $I$ is finite.

In that case let $Y = X \setminus I$. Note that if $y \in Y$, every open nbhd $V$ of $y$ is infinite. Now build the desired discrete set recursively. Start by choosing distinct points $y_0,y_1 \in Y$, and let $W_0,V_1$ be disjoint open sets with $y_0 \in W_0$ and $y_1 \in V_1$. (Here there's a reason for using different letters: the $V_n$'s are needed temporarily for the construction, but it's the $W_n$'s that we really want.) $V_1 \cap Y$ is infinite (why?), so we can pick a point $y_2 \in V_1 \setminus \{y_1\}$. Now let $W_1$ and $V_2$ be disjoint open subsets of $V_1$ such that $y_1 \in W_1$ and $y_2 \in V_2$. Now $V_2 \cap Y$ is infinite, so we can pick a point $y_3 \in V_2 \setminus \{y_2\}$ and let $W_2$ and $V_3$ be disjoint open subsets of $V_2$ such that $y_2 \in W_2$ and $y_3 \in V_3$. Continue in this fashion to get points $y_n$ and open sets $W_n$ for each $n \in \omega$. Clearly $y_n \in W_n$ for each $n$, and with a little thought you should be able to see that if $m \ne n$, $y_m \notin W_n$. (It's probably best to consider the cases $m<n$ and $m>n$ separately. It may also help to make a sketch of the first few steps of the construction.)

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a) Hausdorff implies $T_1$ which implies singletons are closed. Note that closure of $\{x\}$ is defined: $\bigcap\{F:x\in F\}$, and this must equal $\{x\}$ by the aforementioned property of $T_1$ spaces.

b) Choose a point $y \not= x$. We note that there is an open set containing $x$, that does not contain $y$ by choosing 2 open sets, one containing $x$, the other containing $y$, such that the sets are disjoint. The result follows - if $y$ is not in one open set containing $x$, it is certainly not in every open set containing $x$.

c) a: Use the cofinite topology on the set of reals. Singletons are again closed. The fact that every open set intersects every other open set (except $\emptyset$) shows that this topology is not Hausdorff.

b: Use the cofinite topology on the reals. Consider open sets of the form $\{r\}^c$ where $r$ is a real number different from $\{x\}$. The intersection of all these sets is {x} but again, the space is not Hausdorff.

Part 2: Edit: This construction doesn't work, I can't think of a way to fix it sorry. Let $H$ be a Hausdorff space. Let's construct the discrete space inductively.

Induction

Suppose we have the set of points $\{x_i : 1\le i \le n\}$ and a corresponding collection of open sets $O_i$ such that
1.$x_i \in O_i$
2. $x_j \not \in O_i$ where $j \not= i$
3. $\bigcup_{i=1} ^n O_i$ does not cover $H$.

Choose points $x_{n+1}$ and $y$ in $[\bigcup_{i=1}^n O_i]^c$. Choose an open set $O_{n+1}$ around $x_{n+1}$ that does not contain $y$ or $x_i : 1\le i \le n$. (We can find this set by intersecting $n+1$ open sets that contain $x_{n+1}$ which are disjoint from the $x_i : 1\le i \le n$, and $y$. Now we have set of points $\{x_i : 1\le i \le n+1\}$ and a corresponding collection of open sets $O_i$ such that
1. $x_i \in O_i$
2. $x_j \not \in O_i$ where $j \not= i$
3. $\bigcup_{i=1} ^{n+1} O_i$ does not cover $H$.

Base Case

Chose $x_1$ and $y$ in $H$, and choose $O_1$ which contains $x_1$ but not $y$.

Conclusion

Now we have shown the existence of an increasing sequence of subspaces $S_n = \{x_i: 0 \le i \le n\}$ which are discrete. Just consider that $\{x_i\} = S_n \cap O_i \Rightarrow \{x_i\}$ is open in the subspace topology. (By increasing sequence I mean $S_n \subset S_{n+1}$) It remains to be shown that the limit is discrete.

Let $S_{\infty} = \bigcup_n S_n = \{x_i : 1\le i\}$. Choose any point $x_i$ in $S_{\infty}$. Note that, by construction, $O_i \cap S_{\infty} = \{x_i\}$ so all singletons are open and therefore $S_{\infty}$ is a discrete subspace which is countable and infinite.

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why must it be discrete? –  user10 Jul 7 '11 at 6:30
@Mark: the countable set need not be discrete, consider $\{0\}\cup\{\frac{1}{n}\mid n\in\omega\}$. It contains a discrete set, but $0$ is not isolated, therefore the set itself is indiscrete. –  Asaf Karagila Jul 7 '11 at 6:34
@Asaf Karagila: please see emis.de/journals/HOA/IJMMS/6/197.pdf , page $1$, lemma $1$. –  user10 Jul 7 '11 at 6:38
@user10: I did not say it wasn't true. I said that just taking a countable subset does not guarantee that it is discrete. –  Asaf Karagila Jul 7 '11 at 6:40
@Mark: Your construction for (2) doesn't quite work: it's possible that at stage $n$ the open sets $O_0,\dots,O_n$ already cover $X$, in which case you can't choose $x_{n+1}$. Even if they don't cover $X$, their union may be dense in $X$, and you won't be able to choose $O_{n+1}$ disjoint from the $O_i$ with $i<n$. –  Brian M. Scott Jul 7 '11 at 8:16

Your answer to 1(a) is too complicated. Note that $\{x\}$ is in the collection whereof you are taking the intersection.

Also, your answer to 1(c) is too complicated: The natural numbers under the cofinite topology provides a counterexample.

For 2, I haven't thought through the details, but I think this works: First show that any infinite Hausdorff space must have a nonempty open set whose complement is infinite. Then apply the Axiom of Choice inductively to carve out neighborhoods of points.

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