14
$\begingroup$

Let $X$ be an infinite set. Are there Hausdorff topologies $\tau_1, \tau_2$ on $X$ such that $\tau_1\cap\tau_2 = \{\emptyset\} \cup \{U\subseteq X: X\setminus U\text{ is finite}\}$?

(That is, the intersection is as small as it can get.)

And what about the special case $X=\mathbb{R}$ and $\tau_1$ being the topology coming from the Euclidean metric?

$\endgroup$
13
  • $\begingroup$ Cool - can you jot down an example as an answer? $\endgroup$ Mar 22, 2016 at 10:24
  • 3
    $\begingroup$ Every $T_1$ topology contains every cofinite set, so being "as disjoint as possible" means at least including all of these. $\endgroup$
    – Will Brian
    Mar 22, 2016 at 10:48
  • $\begingroup$ Oh right - thanks for this hint! Will edit the question accordingly. $\endgroup$ Mar 22, 2016 at 10:54
  • 8
    $\begingroup$ If $\tau_1$ and $\tau_2$ meet your requirement, then they are called $T_1$-independent. The following paper may contain useful information: Shakhmatov, D.; Tkachenko, M.; Wilson, R. G. Transversal and T1-independent topologies. Houston J. Math. 30 (2004), no. 2, 421–433. $\endgroup$ Mar 22, 2016 at 13:13
  • 1
    $\begingroup$ The paper mentioned by Lajos answers your first question: the authors show that there are two topologies on a set of size $2^{2^{\aleph_0}}$ (both are realizations of the space $\omega^* = \beta \omega - \omega$) whose intersection is precisely the cofinite sets. They don't seem to answer your second question, but there are some relevant-looking results, so you'll want to take a look. $\endgroup$
    – Will Brian
    Mar 22, 2016 at 14:05

3 Answers 3

9
$\begingroup$

Let $\varepsilon$ denote the Euclidean topology on $\mathbb R$.

Proposition: There is a 0-dimensional $T_2$ topology $\tau$ on $\mathbb R$ such that $\tau$ and $\varepsilon$ intersect in only the co-finite sets.

Proof: We need the following lemma:

Lemma: Let $Y=(Y,\nu)$ be an infinite topological space such that $$|\overline{E}^\nu|\ge 2^{\omega}$$ for all $E\in [Y]^\omega$. Then there is an injective map $f:\mathbb R\to Y$ such that $$ \forall D\in [\mathbb R]^\omega \text{ if }\overline D\ne \mathbb R \text{ then } \exists x_D\in (\mathbb R\setminus \overline D) \ f(x_D)\in \overline{f[D]}^\nu. $$

Proof of the lemma:
Enumerate $[\mathbb R]^\omega$ as $\{D_{\alpha}:{\alpha}<2^{\omega}\}$.

By transfinite induction define an increasing continuous sequence $(f_{\alpha}:{\alpha}\le 2^{\omega})$ of injective functions from subsets of $\mathbb R$ into $Y$ such that $|f_{\alpha}|\le {\alpha}+{\omega}$, $D_{{\beta}}\subset dom(f_{\alpha})$ for ${\beta}<{\alpha}$, and $$\text{ if ${\beta}<{\alpha}$ and }\overline D_{\beta}\ne \mathbb R \text{ then } \exists x_{\beta}\in (dom(f_{\alpha})\setminus \overline D_{\beta}) \ f_{\alpha}(x_{\beta})\in \overline{f_{\alpha}[D_{\beta}]}^\nu. $$

Assume that ${\alpha}={\beta}+1$, and we have constructed $f_{\beta}$. Let $g\supset f_{\beta}$ be an injective function from $dom(f_{\beta})\cup D_{\beta}$ into $Y$.

If $\overline{D_{\beta}}=\mathbb R$, then$f_{\alpha}=g$ works.

Assume that $U=\mathbb R\setminus \overline{D_{\beta}}\ne \emptyset$. Since $|U|=2^{\omega}$ and $|\overline{g[D_{\beta}]}^{\nu}|\ge 2^{\omega}$ we can pick $x_{\beta}\in U\setminus dom(g)$ and $y_{\beta}\in \overline{g[D_{\beta}]}^{\nu}\setminus ran(g)$. Let $$f_{\alpha}=g\cup\{(x_{\beta},y_{\beta})\}$$

This completes the inductive construction. QED.

Pick a 0-dimensional $T_2$ space $(Y,\nu)$ which meets the requirements of the lemma. (For example, $Y=\omega^*$ works)

Apply the lemma to obtain an injective $f:\mathbb R\to Y$.

Define the topology $\tau$ by declaring that $f$ is a homeomorphism between $(\mathbb R,\tau)$ and $(f[\mathbb R],{\nu})$.

To show that $\tau $ is as required assume that $\emptyset\ne U\in \varepsilon $ such that $F=\mathbb R\setminus U$ is infinite. Let $D$ be a countable $\varepsilon$-dense subset of $F$. Then there is $x_D\in \mathbb R\setminus \overline D$ such that
$f(x_D)\in\overline{f[D]}^{\nu}$. Since $f$ is a homeomorphism, $x_D\in\overline{D}^{\tau}$.

Thus $D\subset F$ implies that $x_D\in \overline{F}^{\tau}\setminus F$, and so $F$ is not $\tau$-closed. Thus $U\notin\tau$.

Thus we proved the proposition.

$\endgroup$
6
  • $\begingroup$ Very nice! And this argument seems to apply to lots of other familiar spaces as well (any perfect Polish space, for example, or, generalizing from $\omega^*$ to $U(\kappa)$, any space of the form $[0,1]^\kappa$ or $2^\kappa$, if $\kappa$ is regular). $\endgroup$
    – Will Brian
    Mar 23, 2016 at 17:28
  • 1
    $\begingroup$ Wonderful, thanks a lot Lajos...! Intuitively speaking, could it be that since the Euclidean topology $\varepsilon$ is "very connected" (in a quite vague sense of the word), a $T_2$-topology that is "as disjoint as it gets" from $\varepsilon$ must be "very disconnected" (such as the one you constructed? But maybe my intuition is just rubbish :-) $\endgroup$ Mar 24, 2016 at 7:45
  • $\begingroup$ I realise that having asked 2 questions that each get an answer in a different post puts me in the difficult position to select the "canonical" accepted answer. I decided that since Lajos found the paper that Will's answer is based on, and since Lajos put in a lot of work to prove the result in this post, he deserves to get his answer the accepted one, so I changed acceptance. Will - I hope this doesn't annoy you too much! $\endgroup$ Mar 24, 2016 at 7:50
  • $\begingroup$ Dominic, I can not see that connectedness is important here because I think that using my argument one can prove the following statement: If $X=(X,\rho)$ s.t. $$z(X)< |X|=|\rho|=min\{|U|:U\in \rho, U\ne \emptyset \}\le 2^{2^\omega}$$ then there is a topology $\tau$ on $X$ s.t $\rho$ and $\tau $ intersect in only the co-finite sets. $\endgroup$ Mar 25, 2016 at 11:08
  • $\begingroup$ @DominicvanderZypen: No worries -- I agree that this answer deserves to be accepted. $\endgroup$
    – Will Brian
    Mar 29, 2016 at 18:45
13
$\begingroup$

In the comments, Lajos Soukup pointed out the following paper, which seems to contain lots of relevant information:

D. Shakhmatov, M. Tkachenko, and R. G. Wilson, "Transversal and $T_1$-independent topologies," Houston Journal of Mathematics vol. 30, no. 2 (2004), pp. 421 - 433.

(I will give references to the paper below, but I will paraphrase the results to avoid defining all of their terminology.)

They answer Dominic's first question in the negative:

Proposition 3.2: If $X$ is countably infinite, then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.

They are also able to strengthen this result under various extra hypotheses:

Theorem 3.3: It is consistent (it follows from $\neg$CH plus a fragment of MA) that if $|X| < \mathfrak{c}$ then no two Hausdorff topologies on $X$ intersect in only the co-finite sets.

Corollary 3.5: If $|X| < 2^{\aleph_1}$, then no two compact Hausdorff topologies on $X$ intersect in only the co-finite sets.

On the other hand, they show that Dominic's question can have a positive answer for some sets of larger cardinality:

Corollary 3.8: Let $|X| = 2^{2^{\aleph_0}}$. There are two topologies on $X$, both of which realize the compact Hausdorff space $\beta \omega - \omega$, such that the intersection of these two topologies is the set of co-finite subsets of $X$.

See also the preceding lemma, which is more general. (The proof seems to rely heavily on the fact that infinite closed sets have full cardinality, so the technique does not seem to apply to Dominic's second question.)

As for Dominic's second question, I can't seem to find an answer anywhere. But the authors of the above paper do still give us something:

Corollary to Proposition 3.1: If $\tau$ is a Hausdorff topology on $\mathbb{R}$ such that the only sets common to $\tau$ and the usual topology on $\mathbb{R}$ are the co-finite sets, then $\tau$ is countably compact and contains no non-trivial convergent sequences.

I was able to access the Shakhmatov-Tkachenko-Wilson paper here, and I also found some very interesting related information here, on some slides from a talk of Blaszczyk.

$\endgroup$
1
  • $\begingroup$ Beautiful, thanks Will, also the comments concerning $\mathbb{R}$! $\endgroup$ Mar 22, 2016 at 19:47
10
$\begingroup$

If $\tau_1$ and $\tau_2$ meet your requirement, then they are called $T_1$-independent. The following paper may contain useful information: Shakhmatov, D.; Tkachenko, M.; Wilson, R. G. Transversal and $T_1$-independent topologies. Houston J. Math. 30 (2004), no. 2, 421–433. MR2852951.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.