Metric is jointly continuous: Difference between revisions
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==Statement== | ==Statement== | ||
Let <math>(X,d)</math> be a [[metric space]]. Then <math>X</math> is also a topological space in the induced topology, and we can consider the metric as a map of topological spaces <math>d:X \times X \to \R</math>. This map is jointly continuous, i.e. it is continuous from <math>X \times X</math> given the product topology. | Let <math>(X,d)</math> be a [[metric space]]. Then <math>X</math> is also a [[topological space]] in the induced topology, and we can consider the metric as a map of topological spaces <math>d:X \times X \to \R</math>. This map is jointly continuous, i.e. it is continuous from <math>X \times X</math> given the product topology. | ||
==Definitions used== | |||
===Metric space=== | |||
===Topology induced by a metric=== | |||
===Product topology=== | |||
===Continuous map=== | |||
==Proof== | ==Proof== | ||
It suffices to show that inverse images of open subsets of the form <math>(-\infty,a)</math> and <math>(b,\infty)</math> are open subsets of <math>X \times X</math>. We will use the triangle inequality to prove this. | It suffices to show that inverse images of open subsets of the form <math>(-\infty,a)</math> and <math>(b,\infty)</math> are open subsets of <math>X \times X</math>. We will use the triangle inequality to prove this. | ||
Latest revision as of 21:25, 19 July 2008
Statement
Let be a metric space. Then is also a topological space in the induced topology, and we can consider the metric as a map of topological spaces . This map is jointly continuous, i.e. it is continuous from given the product topology.
Definitions used
Metric space
Topology induced by a metric
Product topology
Continuous map
Proof
It suffices to show that inverse images of open subsets of the form and are open subsets of . We will use the triangle inequality to prove this.