First countable space and separability question

  • Thread starter Thread starter radou
  • Start date Start date
  • Tags Tags
    Space
Click For Summary
SUMMARY

The discussion addresses two assignments regarding topological spaces: (Z, P(Z)) and (N, U). It concludes that (Z, P(Z)) is separable because it is a countable space, making itself a dense subset. Additionally, (N, U) is determined to be first-countable since each point has a countable neighborhood basis derived from the open sets Oi. Both solutions are affirmed as correct, emphasizing that the reasoning behind them is straightforward.

PREREQUISITES
  • Understanding of topological spaces and their properties
  • Familiarity with countable sets and dense subsets
  • Knowledge of neighborhood bases in topology
  • Basic concepts of first-countable and second-countable spaces
NEXT STEPS
  • Study the properties of separable spaces in topology
  • Learn about countable dense subsets and their implications
  • Explore first-countable and second-countable spaces in detail
  • Investigate the role of neighborhood bases in topological spaces
USEFUL FOR

Students and researchers in mathematics, particularly those focusing on topology, as well as educators looking for clear examples of separability and first-countability in topological spaces.

radou
Homework Helper
Messages
3,149
Reaction score
8

Homework Statement



Here are two assignments the solutions of which seem somewhat too easy, so I want to check them. :)

1) Is the topological space (Z, P(Z)), where Z is the set of integers, and P(Z) its partitive set, separable?

2) Is the topological space (N, U), where U = {0}U{Oi : i is in N}, Oi = {i, i+1, i+2, ...}, a first-countable space? ("0" denotes the empty set)

The Attempt at a Solution



1) It seems to me that the only set which is dense in Z is Z itself, since every element of P(Z) intersects it. Since, Z is countable, we conclude that (Z, P(Z)) is separable.

2) A space is first-countable if each point has a countable neighbourhood basis. Now, if i is some natural number, then the neighbourhoods of i are Oi, Oi-1 , ... , O1. This family is also a neighbourhood basis for i, since, if U is any neighbourhood of i, there exist one or more sets from the basis such that Oi is contained in U. Since the number of these sets is always countable, we conclude that (N, U) is first countable.
 
Physics news on Phys.org
Your solutions to both parts are correct, but for both parts you are thinking too hard.

For part 1: A countable space is always separable for the simple reason that the space itself is always a countable dense subset.

For part 2: A space with only countably many distinct open sets is always first countable (in fact second countable) because every neighborhood basis at a point (or every basis for the topology) is contained in the (countable) topology.
 
ystael said:
Your solutions to both parts are correct, but for both parts you are thinking too hard.

For part 1: A countable space is always separable for the simple reason that the space itself is always a countable dense subset.

For part 2: A space with only countably many distinct open sets is always first countable (in fact second countable) because every neighborhood basis at a point (or every basis for the topology) is contained in the (countable) topology.

Thanks, I have thought about these facts, they're simple and obvious, but I'm a beginner here so I'm cautious. :)
 

Similar threads

Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K