Metric and Topological Spaces

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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.

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


  1. In the real line R an open neighbourhood is the open interval (p - epsilon, p + epsilon).
  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 dinfinity 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 (xi) rarrow x in a metric space if every epsilon-neighbourhood contains all but a finite number of terms of (xi).
  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.

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.


  1. An open interval (0, 1) is an open set in R with its usual metric.
    Choose epsilon < min {a, 1-a}. Then Vepsilon(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 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 iso (0, 1) cross (0, 1) ].
    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) belongs 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 noteq y) every subset is open.

Properties of open sets
  1. The union (of an arbitrary number) of open sets is open.
    Let x belongs Ai = A. Then s belongs Ai for some i. Since this is open, x has an epsilon-neighbourhood lying completely inside Ai and this is also inside A.

  2. The intersection of finitely many open sets is open.
    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 epsilon1-neighbourhood lying in A. Similarly x has an epsilon2-neighbourhood lying in B. So if epsilon = min {epsilon1 , epsilon2} 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 {epsilon1 , epsilon2 , epsilon3 , ...} > 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.

    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 belongs X | f(x) belongs B} still exists.
    Take x belongs f -1(B). Then f(x) = y belongs B.
    Since B is open the point y has an epsilon-neighbourhood subset B.
    Then, from the definition of continuity this epsilon-neihbourhood contains the image of some delta-neighbourhood V of x. Since f(V) subset B we have V subset f -1(B) and so x has this delta neighbourhood subset f -1(B).
    Hence f -1(B) is open in X.

    The converse of this result also holds:

    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.

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

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.

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(Limit points and closed sets in metric spaces)

JOC February 2004