Is the Sorgenfrey line second-countable?
The “half-open” intervals of the form [a,b) form a basis for a certain topology on R. This topological space is known as the Sorgenfrey line. The rational numbers are a dense subset of this topological space, so the Sorgenfrey line is separable. The Sorgenfrey line is not second countable.
Is the real line second-countable?
The real line is the second countable space. Any uncountable set X with a countable topology is not the first countable and so is not the second countable space.
Is lower limit topology second-countable?
R with the lower limit topology is not second-countable.
Is R with lower limit topology first countable?
Consider Rl, X = R with the lower limit topology which has basis {[a, b) | a < b,a,b ∈ R}. Given x ∈ Rl, the set of all basis elements of the form {[x, x + 1/n) | n ∈ N} is a countable basis at x and so Rl is first-countable. Also, the rationals Q are dense in Rl, so Rl is separable.
Why lower limit topology is not Metrizable?
The real line with the lower limit topology is not metrizable. The usual distance function is not a metric on this space because the topology it determines is the usual topology, not the lower limit topology. This space is Hausdorff, paracompact and first countable.
Is Hilbert space second-countable?
Hilbert Sequence Space is Second-Countable.
Are hausdorff spaces second-countable?
A Hausdorff locally Euclidean space is second-countable precisely it is paracompact and has a countable set of connected components. In this case it is called a topological manifold.
Is second-countable a topological property?
Second-countability implies certain other topological properties. Specifically, every second-countable space is separable (has a countable dense subset) and Lindelöf (every open cover has a countable subcover). The reverse implications do not hold.
What is a second-countable topological space?
A topological space is second-countable if it has a base for its topology consisting of a countable set of subsets. Definition 2.2. A locale is second-countable if there is a countable set B of open subspaces (elements of the frame of opens) such that every open G is a join of some subset of B.
Is Sorgenfrey a compact line?
We observe also that the Sorgenfrey line cannot be compact (since the usual topology on R is coarser and not compact). However, the Sorgenfrey line is hereditarily Lindelöf, i.e. every subspace is Lindelöf (Exercise).
Is the cofinite topology second countable?
(xxi) The space N with the cofinite topology is second countable. True. Hint: There are only countably many cofinite subsets of N. (xxii) Let X be any set endowed with the cofinite topology.
Is every first countable space metrizable?
For every open set containing a ball exists. Hence every metric space is a first countable space.
Is Hilbert space countable?
A Hilbert space H is separable (that is, has a countable dense subset) if and only if it has one countable orthonormal basis if and only if every orthonormal basis for H is countable.
Is every first countable space is second-countable?
A second countable space has a countable basis B − which consist of a countable family of open sets − then the members of B which contain a particular point a form a countable local basis at a. Thus each second countable space is first countable.
Is the cofinite topology second-countable?
Is Sorgenfrey line normal?
Next, we observe that the Sorgenfrey line is normal since it is both regular and Lindelöf. Finally, we prove that the Sorgenfrey plane is not normal, and hence the product of two normal spaces need not be normal.
Is the Sorgenfrey plane separable?
The Sorgenfrey plane is also an example of a separable space, but admitting an uncountable discrete subspace (which is thus not separable), showing that separability is not a hereditary property.
Is cofinite topology 1st countable?
The set of real numbers with the cofinite topology is not a first countable space. The cofinite topology on a set contains and the complements of finite sets. Suppose is a first countable space. Let be a countable open base at Each is open so is closed hence finite.
Is cofinite topology is separable?
topology on X. Z =⋂{C ⊆ R : Z ⊆ C, C is closed in R with the given topology} = R, since the only closed set in R containing Z is R. Thus, R with finite complement topology is separable because it contains a countable dense subset.