If F is Continuous in the Usual Metric Then Show That There Exists an Open Neighborhood

Metric and Topological Spaces


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) epsilon-neighbourhood of a point p is the set of all points within epsilon of it.

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

Examples

  1. In the real line R an open neighbourhood is the open interval (p - epsilon, p + epsilon).
  2. In R 2 (with the usual metric d 2) an open neighbourhood is an "open disc" (one not containing its boundary); in R 3 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 infinity one can recognise an epsilon-neighbourhood of a point (or function) f as the set of functions whose graphs lie in an epsilon-band around the graph of f.

Use of the idea of neighbourhood allows us the rephrase our important analytic definitions:
  1. A sequence (x i ) rarrow x in a metric space if every epsilon-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 belongs X if every epsilon-neighbourhood of f(p) contains the image of some delta-neighbourhood of p.

We can now use the concept of an epsilon-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 epsilon-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 epsilon < min {a, 1-a}. Then V epsilon (a) subset (0, 1).

    Note, however, that there are other subsets of R which are open but which are not open intervals. For example (0, 1) union (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) belongs R 2 | x 2 + y 2 < 1} is an open subset of R 2 with its usual metric.
    So also is "an open square" [a square without its boundary iso (0, 1) cross (0, 1) ].
    Proof
    Any point can be in included in a "small disc" inside the square.

    In general, any region of R 2 given by an inequality of the form {(x, y) belongs R 2 | 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 noteq y) every subset is open.

Properties of open sets
  1. The union (of an arbitrary number) of open sets is open.
    Proof
    Let x belongs A i = A. Then s belongs A i for some i. Since this is open, x has an epsilon-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 belongs A intersect B.
    Then x belongs A and so has an epsilon 1-neighbourhood lying in A. Similarly x has an epsilon 2-neighbourhood lying in B. So if epsilon = min {epsilon 1 , epsilon 2} this epsilon-neighbourhood lies in both A and B and hence in A intersect 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 {epsilon 1 , epsilon 2 , epsilon 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 rarrow Y is a continuous function between metric spaces and B subset Y is open, then f -1(B) is an open subset of X.

    If f:X rarrow 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.

Remark
We will see later that the above properties of open sets in metric spaces are the basis for the generalisation from metric to topological spaces.


JOC February 2004

barnesacyll1993.blogspot.com

Source: http://www-groups.mcs.st-andrews.ac.uk/~john/MT4522/Lectures/L8.html

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