Limit points and closed sets in metric spaces

Definition
If A is a subset of a metric space X then x is a limit point of A if it is the limit of an eventually non-constant sequence (a_i) of points of A.

Remarks

1. This is the most common version of the definition -- though there are others.
   
2. Limit points are also called accumulation points.
   

1. R with the usual metric
   Sets sometimes contain their limit points and sometimes do not.
   
   1. The points 0 and 1 are both limit points of the interval (0, 1).
      
   2. The set Z R has no limit points.
      For example, any sequence in Z converging to 0 is eventually constant.
      
   
   
2. R² with the usual metric
   The limit points of the open disc {(x, y) R² | x²+ y² < 1} form the closed disc {(x, y) R² | x²+ y² 1}.
   Any point on the boundary of the circle is a limit of a sequence of points inside the circle.

3. In R every real number is a limit point of the subset Q of rationals.
   Proof
   Every real number can be approximated arbitrarily closely by a sequence of rationals (by truncating the decimal expansion, say).
   
   
   

1. A point x of X is a limit point of A if every -nd of x in X meets A in a point x.
   
2. A point x of X is a limit point of A if every open set of X containing x meets A in a point x.
   

Examples

1. The subset X is a closed subset of itself. The empty set is closed. Any finite set is closed.
   
2. The closed interval [0, 1] is closed subset of R with its usual metric. There are, however, lots of closed subsets of R which are not closed intervals.
   
3. The closed disc, closed square, etc. are closed subsets of R².
   

Theorem
A subset A of a metric space X is closed if and only if its complement X - A is open.

Proof
() Suppose A is closed. We need to show that X - A is open. So suppose that x X - A. Then some -neighbourhood of x does not meet A (otherwise x would be a limit point of A and hence in A). Thus this -neighbourhood of x lies completely in X - A which is what we needed to prove.

() Conversely, suppose that X - A is open. We need to show that A contains all its limit points. So suppose x is a limit point of A and that x A. Then x X - A and hence has an -neighbourhood X - A. But this is an -neighbourhood that does not meet A and we have a contradiction.

Recap
The last two sections have shown how we can phrase the ideas of continuity and convergence purely in terms of open sets.

1. A map f: X Y is continuous B open in Y f ^-1(B) is open in X.
   
2. A point x is a limit point of a subset A of X every open set containing x meets A (in a point x).