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There is quite an easy isomorphism that you may just have overlooked: ${\cal I}(F(\mathbb{N})) \cong {\cal P}(\mathbb{N})$, where ${\cal P}(\mathbb{N})$ denotes the powerset of $\mathbb{N}$. The isomo …

Yes - set $L:=\kappa+2$ and endow it with the following ordering: $\kappa < \alpha$ for all $\alpha \in \kappa$; $\kappa+1 > \alpha$ for all $\alpha \in \kappa$. (Essentially this is an infinite v …

Let $L$ be an complete lattice. A lattice homomorphism $f: L\to L$ is said to be join-incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_L f(S).$ If $f:L\to …

Let ${\cal L}$ be defined as in this question. Is there a surjective lattice homomorphism $f: {\cal L}\to \mathbb{N}^\mathbb{N}$, where $\mathbb{N}^\mathbb{N}$ is the set of all functions, ordered poi …

Let $L$ be a complete lattice with top element $1$. Let ${\cal M}(L)$ be the collection of meet-irreducible elements of $L$. Recall that a frame is said to be spatial if for all $x\in L$ with $x<1$ w …

The answer is yes, and a key ingredient is the following theorem due to George Grätzer: A bounded distributive lattice is affine complete if and only i f it does not contain a proper …

If we consider $\omega^\omega$ as a lattice with component-wise join and meet, is there a finite distributive lattice $L$ so that there is no injective lattice homomorphism $f:L\to\omega^\omega$?

Let $L$ be a lattice and let $\textbf{DM}(\cdot)$ denote the Dedekind-MacNeille completion. Is there a lattice $L$ that is not a quotient of $\textbf{DM}(L)$? And what if we generalise this question …

Let $L, K$ be complete lattices. A lattice homomorphism $f: L\to K$ is said to be incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_K f(S).$ Consider the f …

Let $L, K$ be complete lattices. A lattice homomorphism $f: L\to K$ is said to be incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_K f(S).$ Suppose that $ …

Let $L$ be an complete lattice. A lattice homomorphism $f: L\to L$ is said to be join-incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_L f(S).$ Suppose $L …

For any lattice $L$ we denote the complete lattice of the ideals of $L$ by ${\cal Id}(L)$. If $L$ is complete, is there a lattice homomorphism from ${\cal Id}(L)$ onto $L$?

Inspired by Keith's example, let me write down an easier to understand (but essentially the same) lattice: Let $L = \omega^{<\omega}\cup \{\omega\}$ where $\omega^{<\omega}$ denotes the set of finite …

Let $L$ be an complete lattice. A lattice homomorphism $f: L\to L$ is said to be join-incomplete if there is an infinite set $S \subseteq L$ such that $f(\bigvee_L S) \neq \bigvee_L f(S).$ Is the fol …

No, $T_2$-spaces with this property aren't necessarily homeomorphic. Let $\mathbb{N} = \{0,1,2,\ldots\}$ be the set of non-negative integers and let $\tau = {\cal P}(\mathbb{N})$ be the discrete topo …