Metric Space, open and closed sets

[Ankit Mishra]

1. The problem statement, all variables and given/known data

Let X be set donoted by the discrete metrics
d(x; y) =(0 if x = y;
1 if x not equal y:
(a) Show that any sub set Y of X is open in X
(b) Show that any sub set Y of X is closed in y

2. Relevant equations

In a topological space, a set is closed if and only if it coincides with its closure. Equivalently, a set is closed if and only if it contains all of its limit points.

A subset U of a metric space (M, d) is called open if, given any point x in U, there exists a real number ε > 0 such that, given any point y in M with d(x, y) < ε, y also belongs to U. Equivalently, U is open if every point in U has a neighbourhood contained in U.

3. The attempt at a solution

I tried to use the line y=x and said that there is a finite many balls centered along the line, that means its open because of the union. But my prof said that we r working in abstract so we cant use R2. How would this work then?

[VeeEight]

For (b), it is clear that if we fix a point x, the set {y | d(x,y) < 1/2} is just the point x What are the limit points of X?

For (a), use the fact that a set is open if it's complement is closed. If all the subsets of X are closed, then the complement of every subset is closed.

[Ankit Mishra]

(b) Show that any sub set Y of X is closed in X.....is the correct question

[VeeEight]

Yes, I assumed that's what was meant.

[Ankit Mishra]

the set {y | d(x,y) < 1/2}.....how was developed? and why less than 1/2?

[lanedance]

there are only really two cases here as the distance between every point is 1, the two cases are

d>=1
d<1
consisder what points aere within a ball of each radius in each case

the choice of d<1/2 was probably an abritray d<1 case to consider

[lanedance]

this is known as the discrete topology of X