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A poset or partially ordered set is a set endowed with a partial order, meaning a binary relation $\leq$ which is reflexive ($x \leq x$ for all $x$), antisymmetric ($x\leq y$ and $y\leq x$ implies $x=y$), and transitive ($x\leq y$ and $y\leq z$ implies $x \leq z$).
6
votes
Accepted
$\text{Max}\big(\text{Sub}(\text{Sym}(\omega))\setminus \{\text{Sym}(\omega)\}\big)$
As was written by Andreas Thom in his comment, Google gives an extensive literature on this subject. For example, look at the papers
https://link.springer.com/chapter/10.1007/978-94-011-2080-7_18
or h …
4
votes
Accepted
Order convergence vs topological convergence in partially ordered sets
The answer to this question is negative.
As a counterexample, consider the one-point extension $P:=2^{<\omega}\cup\{\infty\}$ of the binary tree $2^{<\omega}=\bigcup_{n\in\omega}2^n$. Here $2$ is the …
4
votes
Accepted
Topologies with no minimal $T_2$ topologies above them
There exists a topology $\tau$ with two non-isolated points on a countable set $X$ such that the poset $T_2(\tau)$ does not have minimal elements.
To construct such topology $\tau$, take any Hausdorf …
3
votes
Accepted
Minimal Hausdorff topologies compatible with a bunch of functions
The answer to this question is affirmative:
Theorem. There exists a countable set $X$ and an uncountable family $\mathcal F$ of self-functions of $X$ such that the poset $T_2(\mathcal F)$ has no min …
2
votes
Accepted
Spaces that are invariant under some contractions
For a countable space $X$ the discussed property means that $X$ is homeomorphic to $X/A$ for any finite subset $A\subset X$. Let us call this property quotient-homogeneous.
There are many quotient-h …
2
votes
Accepted
Maximal elements in the partially ordered set of image spaces
A positive answer to this problem is given by the known answers to the following problem posed by de Groot in New Scottish book.
Problem 393 (de Groot; 28 May, 1958). Does there exist a (plane) conti …
2
votes
Accepted
Is there some characterization of $\omega^\omega$-base related to $S_\omega$?
This question has negative answer.
To construct a counterexample, consider the set $X=(\omega\times\omega)\cup\{\infty\}$ where $\infty\notin(\omega\times\omega)$ is any point. Then set $X$ is endowe …
1
vote
Accepted
Compactification of order-disconnected spaces
It seems that the answer to this problem is affirmative:
Take a well-behaved pospace $P$. To show that $\beta_2(P)$ is well-behaved, take any clopen upper set $U\subset \beta_2(P)$. We should prove t …
1
vote
Accepted
Thinning directed sets ${\frak P}$ of partitions of $\omega$ with no ${\frak P}$-discrete su...
The answer to this question is negative.
Given a linearly ordered family $\mathfrak C$ of finitary partitions of $\omega$, write $\mathfrak C=\bigcup_{n=1}^\infty\mathfrak C_n$ where $\mathfrak C_n=\{ …