Consider $\mathbb R$ with lower limit topology (generated by taking $[a,b)$ intervals as basis).The topological space thus generated is called Sorgenfrey line.What are some interesting properties of this line in terms of countability axioms and compactness,connectedness?
Topological properties of Sorgenfrey line.
compactnessconnectednessfirst-countablegeneral-topologysecond-countable
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Here's a simple direct proof, which works just as well for the Sorgenfrey topology as for the usual topology of the line.
Let $\mathcal U$ be a collection of Sorgenfrey-open sets that covers $\mathbb R$. Let's say that a set $X\subseteq R$ is countably covered if $X$ is covered by countably many members of $\mathcal U$. We want to show that $\mathbb R$ is countably covered.
Consider any $a\in\mathbb R$, and let $C_a=\{x: x\ge a,\text{ and the interval }[a,x]\text{ is countably covered}\}$. It's easy to see that $\sup C_a=\infty$; assuming the contrary leads to a contradiction. Hence every finite interval $[a,b]$ is countably covered, and so is $\mathbb R=\bigcup_{n\in\mathbb N}[-n,n]$.
P.S. I have been asked to explain why assuming that $\sup C_a=b\in\mathbb R$ leads to a contradiction. Let $b_n=b-\frac{b-a}{2^n}$ for $n=1,2,3,\dots,$ so that $a\lt b_n\lt b$ and $b_n\to b.$ Thus for each $n$ there is a countable collection $\mathcal S_n\subseteq\mathcal U$ such that $[a,b_n]$ is covered by $\mathcal S_n,$ and the half-open interval $[a,b)$ is covered by the countable collection $\bigcup_{n\in\mathbb N}\mathcal S_n.$ Moreover, since $\mathcal U$ covers $\mathbb R,$ there is some $U\in\mathcal U$ such that $b\in U.$ Since $U$ is Sorgenfrey-open, there is some neighborhood $[b,b+\varepsilon)$ of $b$ (with $\varepsilon\gt0$) such that $[b,b+\varepsilon)\subseteq U.$ Then $[a,b+\varepsilon)$ is covered by $\{U\}\cup\bigcup_{n\in\mathbb N}\mathcal S_n,$ whence $b+\frac\varepsilon2\in C_a,$ contradicting our assumption that $b=\sup C_a.$
Your question has two parts:
Are there further topological properties that are invariant under diffeomorphism?
The straightforward answer is this: Diffeomorphisms preserve exactly the same topological properties as homeomorphisms do; nothing more, nothing less.
The reason for this is essentially definitional: A topological property is, by definition, a property that is preserved by homeomorphisms. Since every diffeomorphism is a homeomorphism, every topological property is preserved by diffeomorphisms. And if a particular property of smooth manifolds is preserved by diffeomorphisms but not by homeomorphisms, then it's not a topological property.
The other half of your question was
Are there further non-topological properties that are invariant under diffeomorphism?
This is a more interesting question. But first you have to establish an appropriate category of spaces to work in -- diffeomorphisms are only defined between smooth manifolds, which are topological manifolds with an additional structure called a smooth structure. So the appropriate question to ask is whether there are non-topological properties of smooth manifolds that are preserved by diffeomorphisms. There are, but they're more subtle. For example, one such property for compact smooth manifolds is whether they bound (smoothly) parallelizable manifolds. This is one way that exotic spheres can be distinguished from each other.
Best Answer
The Sorgenfrey plane is an example of a separable space admitting an uncountable discrete subspace (which is thus not separable). Indeed it is hereditarily separable (all subspaces are separable, e.g. the rationals are dense in the whole space) and it is hereditarily Lindelöf, but does not have a countable base.
A proof that it does not admit a countable basis: suppose $B_j$ with $j \in J$ is a base for this space. Then, for every $x \in \mathbb{R}$ there is a $B_{j_x}$ such that $x \in B_{j_x} \subseteq [x, x+1)$. So, if $x$ is different than $y$, say without loss of generality $x < y$, then $x \notin B_{i_y}$ and $ x \in B_{i_x}$, so we get that $B_{i_x} \neq B_{i_y}$. This proves that, since all was arbitrary, we can find as many basis elements as points in $\mathbb{R}$.
Moreover, it is an example of a Hausdorff, perfect, non-metrizable space but First-Countable.
Note that the Sorgenfrey line is also totally disconnected, and you can find a proof here. One can also prove that every compact subspace of the Sorgenfrey line is countable: it is proved very nicely by Henno here.
Here are some other nice properties about it.