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[parent] topology induced by uniform structure (Derivation)

Let $ \mathcal{U}$ be a uniform structure on a set $ X$. We define a subset $ A$ to be open if and only if for each $ x \in A$ there exists an entourage $ U \in \mathcal{U}$ such that whenever $ (x,y) \in U$, then $ y \in A$.

Let us verify that this defines a topology on $ X$.

Clearly, the subsets $ \emptyset$ and $ X$ are open. If $ A$ and $ B$ are two open sets, then for each $ x \in A \cap B$, there exist an entourage $ U$ such that, whenever $ (x,y) \in U$, then $ y \in A$, and an entourage $ V$ such that, whenever $ (x,y) \in V$, then $ y \in B$. Consider the entourage $ U \cap V$: whenever $ (x,y) \in U\cap V$, then $ y \in A \cap B$, hence $ A \cap B$ is open.

Suppose $ \mathcal{F}$ is an arbitrary family of open subsets. For each $ x \in \bigcup \mathcal{F}$, there exists $ A \in \mathcal{F}$ such that $ x \in A$. Let $ U$ be the entourage whose existence is granted by the definition of open set. We have that whenever $ (x,y) \in U$, then $ y \in A$; hence $ y \in \bigcup \mathcal{F}$, which concludes the proof.



"topology induced by uniform structure" is owned by Mathprof. [ full author list (2) | owner history (1) ]
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See Also: uniform neighborhood

Also defines:  uniform topology

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uniformizable space (Definition) by CWoo
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Cross-references: proof, topology, entourage, open, subset, uniform structure
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This is version 4 of topology induced by uniform structure, born on 2002-06-10, modified 2007-02-20.
Object id is 3091, canonical name is TopologyInducedByAUniformStructure.
Accessed 3183 times total.

Classification:
AMS MSC54E15 (General topology :: Spaces with richer structures :: Uniform structures and generalizations)

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