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.

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?

share|improve this question
1  
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 3

up vote 5 down vote accepted

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.)

share|improve this answer

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.

share|improve this answer
    
why must it be discrete? –  user10 Jul 7 '11 at 6:30
3  
@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
1  
@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
1  
@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.

share|improve this answer

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.