Neighbourhoods and open sets in metric spaces

Although it will not be clear for a little while, the next definition represents the first stage of the generalisation from metric to topological spaces.

An (open) -neighbourhood of a point p is the set of all points within of it.

Definition
An open neighbourhood of a point p in a metric space (X, d) is the set V(p) = {x X | d(x, p) < }

Examples

1. In the real line R an open neighbourhood is the open interval (p - , p + ).
   
2. In R² (with the usual metric d₂) an open neighbourhood is an "open disc" (one not containing its boundary); in R³ it is an "open ball" etc.
   
3. Let X be the interval [0, 1] with its usual metric. Then a ¹/₄ -neighbourhood of 0 is the interval [0, ¹/₄).
   Note that this is not an open interval.
   The fact that when one takes a subset of a metric space (called a subspace) the appearance of things like neighbourhoods may change is an important fact that we will need later on.
   
4. In C[0, 1] with the metric d one can recognise an -neighbourhood of a point (or function) f as the set of functions whose graphs lie in an -band around the graph of f.
   

1. A sequence (x_i) x in a metric space if every -neighbourhood contains all but a finite number of terms of (x_i).
   
2. A function f from a metric space X to a metric space Y is continuous at p X if every -neighbourhood of f(p) contains the image of some -neighbourhood of p.
   

We can now use the concept of an -neighbourhood to define one of the most important ideas in a metric space.

Definition
A subset A of a metric space X is called open in X if every point of A has an -neighbourhood which lies completely in A.

Examples

1. An open interval (0, 1) is an open set in R with its usual metric.
   Proof
   Choose < min {a, 1-a}. Then V(a) (0, 1).
   
   
   Note, however, that there are other subsets of R which are open but which are not open intervals. For example (0, 1) (2, 3) is an open set.

2. Let X = [0, 1] with its usual metric (which it inherits from R). Then the subset [0, ¹/₄) is an open subset of X (but not of course of R).
   
3. A set like {(x, y) R² | x² + y² < 1} is an open subset of R² with its usual metric.
   So also is "an open square" [a square without its boundary (0, 1) (0, 1) ].
   Proof
   Any point can be in included in a "small disc" inside the square.
   
   
   In general, any region of R² given by an inequality of the form {(x, y) R² | f(x, y) < 1} with f(x, y) a continuous function, is an open set.
   
4. Any metric space is an open subset of itself. The empty set is an open subset of any metric space.
   We will see later why this is an important fact.
   
5. In a discrete metric space (in which d(x, y) = 1 for every x y) every subset is open.
   

1. The union (of an arbitrary number) of open sets is open.
   Proof
   Let x A_i = A. Then s A_i for some i. Since this is open, x has an -neighbourhood lying completely inside A_i and this is also inside A.
   
   
   
2. The intersection of finitely many open sets is open.
   Proof
   It is enough to show this for just two open sets A and B.
   So suppose x A B.
   Then x A and so has an ₁-neighbourhood lying in A. Similarly x has an ₂-neighbourhood lying in B. So if = min {₁ , ₂} this -neighbourhood lies in both A and B and hence in A B.
   
   
   Note that the same proof will not necessarily work for the intersection of infinitely many sets since we could not be sure that min {₁ , ₂ , ₃ , ...} > 0.
   For example, in R with its usual metric the intersection of open intervals: (-1/i, 1/i) = {0} which is not open.

   We can now connect the concept of continuity with open sets.

3. If f:X Y is a continuous function between metric spaces and B Y is open, then f ^-1(B) is an open subset of X.

   Proof
   Note that even if f is not a one-one-correspondance (and does not have an inverse function f ^-1) the set f ^-1(B) ={x X | f(x) B} still exists.
   Take x f ^-1(B). Then f(x) = y B.
   Since B is open the point y has an -neighbourhood B.
   Then, from the definition of continuity this -neihbourhood contains the image of some -neighbourhood V of x. Since f(V) B we have V f ^-1(B) and so x has this neighbourhood f ^-1(B).
   Hence f ^-1(B) is open in X.
   
   
   The converse of this result also holds:

   If f:X Y is a function for which f ^-1(B) is open in X for every open set B in Y, then f is continuous at every point of X.

   Proof
   To show that f is continuous at x X, take B to be an -neighbourhood of y Y.
   Then f ^-1(B) is open in X and so x has a -neighbourhood in f ^-1(B). This -neighbourhood is mapped inside the -neighbourhood of f(x) and so f is continuous at x.