General Topology/Metric spaces

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Definition (metric):

Let be a set. A metric on is a function such that

  1. ("symmetry")
  2. ("triangle inequality")

Definition (metric space):

Let be a set, together with a metric on it. Then , together with , is called a metric space.

Definition (open ball):

Let be a metric space, . The open ball of size centered at is defined to be the set


Definition (topology of a metric space):

Let be a metric space. The topology of , which is a topology on the set , is defined to be

Failed to parse (unknown function "\middle"): {\displaystyle \left\{ U \subseteq M \middle| \forall x \in U: \exists \epsilon > 0: B_\epsilon(x) \subseteq U \right\}} .

This definition is justified:

Proposition (metric induces topology):

Let be a metric space. Define as follows:

Failed to parse (unknown function "\middle"): {\displaystyle \tau := \left\{ U \subseteq M \middle| \forall x \in U: \exists \epsilon > 0: B_\epsilon(x) \subseteq U \right\}}

Then is a topology on .

Remark (open ball is open):

Note that each open ball is an open subset of the underlying metric space with the topology indicated above, since whenever , we may choose and obtain due to for .

Proof: This topology is precisely the topology that arises in the characterisation of topologies by their neighbourhoods when we choose the neighbourhoods of a point to be those sets that for a sufficiently small contain . It is clear that the neighbourhoods thus defined satisfy 1.-3. of the definition, and by the above remark, they also satisfy 4., since an open ball is a neighbourhood of each of its points.

Now subsets of metric spaces are again metric spaces (as can be easily verified), and the topology induced by the restriction of the metric of the larger space is the subspace topology:

Proposition (metric subspace has subspace topology):

Let be a metric space and be a subset. Then the topology on induced by the metric on , namely , is the subspace topology.

Proof: We only need to prove that they have the same open sets. being open in the subspace topology means that there exists open such that . If that is the case, pick arbitrary. By openness of , choose sufficiently small so that , where the ball is taken with respect to the metric on . Then note that whenever is such that , then because by definition of restrictions, . Therefore, whenever for , we have . For the other direction, suppose that is open in the metric topology induced by on .

Proposition (metric spaces are normal):

Let be a metric space. Then together with the topology induced by the metric is normal.

Proof: Let be two disjoint and closed subsets of . For each , define sufficiently small so that , and similarly for . Then define

, .

Note that , . Suppose , and choose and so that resp. . Both cases and lead to a contradiction by the triangle inequality.

As usual, by choosing , maximal by choosing the union of all the open balls satisfying the condition, one may avoid the axiom of choice.

Proposition (metric spaces are completely normal):

Let be a metric space. Then , together with its subspace topology, is completely normal.

Proof: Let be any subset of . As noted above, has the structure of a metric space, and General Topology/Metric spaces#metric spaces are normal. Further, its subspace topology equals the topology induced by its metric, so that it is normal in the subspace topology. Thus, is hereditary normal, that is, completely normal.

Definition (proper space):

A metric space is called proper if and only if all of its closed balls are compact.

Proposition (closed point projection exists in proper spaces):

Let be a proper metric space, let be closed and let . Then there exists a point which has shortest distance to among all points in .

Proof: Note that by the existence of infima over reals, the value

exists. Now define for

Failed to parse (unknown function "\middle"): {\displaystyle K_n := \left\{z \in A \middle| d(x, z) \le \eta + \frac{1}{n} \right\}} ;

note that for all (since otherwise the infimal distance of to would be larger than ) and further . Further, each is compact as the closed subset of a Hausdorff space. Thus, we may apply Cantor's intersection theorem in order to obtain that

. Thus, let .

Note that for each , by the definition of infimums. Further, for each , by definition of , so that .

Proposition (metric spaces are uniform spaces):

Let be a metric space, and define

Failed to parse (unknown function "\middle"): {\displaystyle U_n :=\left\{ (x, y) \in M \times M \middle| d(x, y) < \frac{1}{n} \right\}} , and then Failed to parse (unknown function "\middle"): {\displaystyle \mathcal U := \left\{ U_n \middle| n \in \mathbb N \right\}} .

Then is the basis for an entourage filter on .

Proof: Clearly, each subset of contains the diagonal. Further, the intersection of any two elements of the basis equals a third due to the computation


this implies that we really have a filter basis. Finally, let be an entourage. Pick such that . Note first that is symmetric, since is symmetric. Then, choose then . Suppose that and , then .

Proposition (metrics may be chosen bounded):

Let be a metric space, and let be the topology on induced by . Then upon defining


we obtain that is a bounded metric that induces the same topology on as .

Proof: It is obvious that for all . Suppose now that . Then


so that becomes equivalent to

proving the triangle inequality, which is the only non-trivial thing that one has to show in showing that is a metric.

Proposition (countable product of metric spaces is metric):

Let be a sequence of metric spaces, and consider the (a priori only topological) space with product topology. Then is metrizable, and a possible metric is


Example (infinite-dimensional real space):

Consider the space , the product of countably many copies of together with the product topology induced by the Euclidean topology on .

Definition (proper space):

A proper space is a metric space so that all its closed balls are compact.

Proposition (existence of closest element in closed subspace for proper spaces):

Let be a metric space with metric , let be closed and a proper space, and let . Then there exists such that


Proof: Set


Then the sets


form a descending family of nonempty compact and closed sets, since closed subsets of compact sets are compact and compact subsets of Hausdorff spaces are closed. Hence, Cantor's intersection theorem proves that there exists


But then for each , , so that in fact


Exercises[edit | edit source]

  1. Let be a compact topological space, whose topology is induced by two metrics and . Prove that for every , there exists such that implies (and then by symmetry also the other way round).