In mathematics, S-space is a regular topological space that is hereditarily separable but is not a Lindelöf space. L-space is a regular topological space that is hereditarily Lindelöf but not separable. A space is separable if it has a countable dense set and hereditarily separable if every subspace is separable.
It had been believed for a long time that S-space problem and L-space problem are dual, i.e. if there is an S-space in some model of set theory then there is an L-space in the same model and vice versa – which is not true.
It was shown in the early 1980s that the existence of S-space...
3
I know that in a complete metric space, if $F_1\supseteq F_2\supseteq...$ are closed, and $\text{diam}(F_n)\to0,$ then $\bigcap\{F_n\}\not=\emptyset.$ John B. Conway claims falsely that even if $\text{diam}(F_N)\not=0,$ but is still bounded for all $n$, then $\bigcap\{F_n\}\not=\emptyset.$ I say ...
Show that a totally bounded complete metric space $X$ is compact.
I can use the fact that sequentially compact $\Leftrightarrow$ compact.
Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$ $\forall\epsilon>0$, $X$ can be covered by a finite number of b...
In number theory, Dirichlet's theorem on Diophantine approximation, also called Dirichlet's approximation theorem, states that for any real numbers
α
{\displaystyle \alpha }
and
N
{\displaystyle N}
, with
1
≤
N
{\displaystyle 1\leq N}
, there exist integers
p
{\displaystyle p}
and
q
{\displaystyle q}
such that
1
≤
q...
Let $\omega$ denote the natural numbers and let $N$ be a countably infinite discrete subset of $\beta \omega \setminus \omega$. If $X = \beta \omega \setminus N$, then $X$ is not locally compact, and every compactification of $X$ is zero-dimensional. That $X$ is not locally compact at any point...
0
This is to finish flawed argument of Anonymous that $X$ has only zero-dimensional compactifications.
I'll prove that $K$ is totally disconnected, which is equivalent since $K$ is compact.
Let $S\subseteq K$ be connected and $|S|\geq 2$ where $K$ is a compactification of $X$. Since $\beta\omega\se...
