# Compact Topology and Coarsest Topology

Let $$(E, \mathcal{T})$$ be a compact Hausdorff space. It is well known that every topology $$\mathcal{U}$$ coarser than $$\mathcal{T}$$ such that $$(E, \mathcal{U})$$ is Hausdorff is equal to $$\mathcal{T}$$.

Is the converse true?

(that is: if $$\mathcal{T}$$ is a coarsest topology amongst Hausdorff topology on $$E$$, then $$(E, \mathcal{T})$$ is compact)

• No. See for example in this paper from A. Smythe and C. A. Wilkins. – martini Mar 27 '12 at 19:41
• you can find the paper by searching "A. Smythe and C. A. Wilkins" on google. That link didn't work for some reason. – Dustan Levenstein Mar 27 '12 at 19:46
• Perhaps this google-Link will do. It does for me, at least. – martini Mar 27 '12 at 19:48
• Your question is probably sufficiently answered by the first comment. But a key word you might look for is "minimal Hausdorff space." Also, note that some spaces don't have "minimal Hausdorff" coarsenings -- the rational numbers for example. – ihaphleas Nov 15 '13 at 14:34

If $$A$$ is the linearly ordered set $$\{1,2,3,...,\omega,...,-3,-2,-1\}$$ with the order topology, and if $$\Bbb Z^+$$ is the set of positive integers with the discrete topology, we define $$X=A\times\Bbb Z^+$$ together with two ideal points $$a$$ and $$-a$$. The topology $$\tau$$ on $$X$$ is determined by the product topology on $$A\times \Bbb Z^+$$ together with the local base

$$M_n^+(a)=\{a\} \cup\{(i, j) \mid i<\omega, j>n\} \text{ and } M_{n}^-(-a)=\{-a\} \cup\{(i, j) \mid i>\omega,j

Visualizing $$X$$:

1. A straightforward consideration of cases shows that $$X$$ is Hausdorff.

2. The collection of all basis neighborhoods form an open covering of $$X$$ with no finite subcovering (Hint: Consider the points $$(\omega,j)$$ for each $$j\in \Bbb Z^+$$). Thus $$X$$ is not compact.

3. $$X$$ is almost compact (i.e. each open cover of the space has a finite subcollection the closures of whose member cover the space). This follows from the fact that the closures of any neighborhoods of $$a$$ and $$-a$$ contain all but finitely many of the points $$(\omega,j)$$. A straightforward consideration of cases shows similarly that the complement of any basis neighborhood is also almost compact.

4. Now suppose $$\tau'\subset\tau$$, suppose $$N$$ is a basis neighborhood of $$\tau$$ and suppose $$\{O_\alpha\}$$ is a $$\tau'$$-open covering of $$X \setminus N$$. It is then a $$\tau$$ open covering, so there exist finitely many sets $$O_1, O_2,..., O_n$$ the union of whose $$\tau$$-closures covers $$X \setminus N$$. But the $$\text{cl}_{\tau} (O_i)\subset\text{cl}_{\tau'} (O_i)$$, so $$X \setminus N$$ is covered by the union of the $$\tau'$$-closures of $$0_1, O_2, ..., O_n$$. In otherwords, $$X\setminus N$$ is an almost compact subset of $$(X,\tau')$$

5. Suppose $$\tau'$$ is a proper subtopology of $$\tau$$. Then there would be some basis neighborhood $$N\in \tau$$ for which $$X\setminus N$$ is not closed in $$\tau'$$, and so there would be a point $$x\in N$$ such that $$x\in \text{cl}_{\tau'} (X\setminus N)$$. Let $$\{C_\alpha\}$$ be the collection of all $$\tau'$$-neighborhoods of $$x$$, and suppose $$\{X\setminus \overline{C_\alpha}\}$$ covers $$X\setminus N$$, then there exists $$\overline{C_1},\overline{C_2},...,\overline{C_n}$$ such that $$X\setminus N\subset \bigcup \left(\overline{X\setminus\overline{C_i}}\right)$$ (by 4). But $$\bigcup\ \left(\overline{X \setminus \overline{C}_{i}}\right)$$ is closed, hence it must contain $$x$$ (as $$x\in \text{cl}_{\tau'} (X\setminus N)$$). But $$\bigcap C_i$$ is a neighborhood of $$x$$, implying $$\left(\bigcap C_{i}\right) \cap\left(\bigcup\ \left(\overline{X \setminus \overline{C}_{i}}\right)\right)=\left(\bigcap C_{i}\right) \cap\left(\overline{\bigcup\ \left(X \setminus \overline{C}_{i}\right)}\right)$$ is nonempty, which is clearly impossible. Hence $$\{X\setminus \overline{C_\alpha}\}$$ doesn't cover $$X\setminus N$$, which implies there exists $$y\notin N$$ such that $$y\in \bigcap \overline{C_\alpha}$$. Since $$y\neq x$$ are inseparable by open sets, $$(X,\tau')$$ is not Hausdorff.

In conclusion, $$X$$ is minimal Hausdorff but not compact.

The above is essentially my paraphrasing of Steen and Seebach's Counterexample in Topology example 100.

• +1... My edit was for a typo in 5. ("contain" for "contains") and to change $X-N$ to $X\setminus N$ at one place in 4. to conform to the other two instances of it in 4.... In 5. I think it would be clearer in 5. to say that $\cap C_i$ is a nbhd of $x$ so if $x\in \cup(\overline {X\setminus \bar C_i})=$ $\overline {\cup (X\setminus \bar C_i)}$ then $\phi\ne (\cap C_i)\cap (\cup (X\setminus \bar C_i)$ which is (more) clearly impossible. – DanielWainfleet May 6 at 21:24
• The space A is $T_2$ as it is linear and $\Bbb Z^+$ is $T_2$ as it is discrete, so $A\times \Bbb Z^+$ is $T_2$. And $A\times \Bbb Z^+$ is open in $X$. So to check that $X$ is $T_2$ we need only to check the effect of adding $a$ and $-a$.....You refer to "the" basis and I know what you mean, but $X$ has many bases. Some readers may be puzzled.... I think the phrasing in 2. is hard to understand...... Altogether, though, a very nice argument. – DanielWainfleet May 6 at 21:36
• @DanielWainfleet Thanks for your kind words! I will edit it if I have time. – YuiTo Cheng May 6 at 23:47