A map is continuous if and only if each point in the domain has an open neighbourhood for which the restriction of the map is continuous on
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Flesh out and demote
- Note: this might be called the local criterion for continuity
Contents
[hide]Statement
Let (X,J) and (Y,K) be topological spaces, and let f:X→Y be a map between them, then[1]:
- f:X→Y is continuous if and only if for each point x∈X there exists an open neighbourhood of x, say U, such that f|U:U→Y[Note 1] is continuous
- Symbolically, the right hand side can be written:
- ∀x∈X∃U∈J[x∈U∧(f|U:U→Y is continuous)][Note 2]
- And the full statement
- Symbolically, the right hand side can be written:
TODO: Symbolical representation of entire statement, including continuity!
Proof
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Continuous ⟹ Statement
Suppose f:X→Y is continuous. We wish to show ∀x∈X∃U∈J[x∈U∧(f|U:U→Y is continuous)]
- Let x∈X be given.
- Then take X∈J as our open neighbourhood of x. Automatically x∈X and f|X:X→Y is just f:X→Y and is continuous, we're done.
Statement ⟹ continuous
Suppose ∀x∈X∃U∈J[x∈U∧(f|U:U→Y is continuous)], we wish to show f:X→Y is continuous.
- Let U∈K be given (so U is open in (Y,K)), we must show f−1(U) is open in (X,J).
- For any/all x∈f−1(U) there exists an open neighbourhood Vx∈J of x such that f|Vx:Vx→Y is continuous, specifically this means:
- (f|Vx)−1(U) is relatively open in Vx (as we must consider Vx with the subspace topology it inherits from (X,J))
- But as Vx is itself open, and the relatively open sets of an open subspace are open[Note 3] we see (f|Vx)−1(U) is open in (X,J) too.
- We see that (f|Vx)−1(U):={x∈Vx | f|Vx(x)∈U} ={x∈Vx | f(x)∈U}=f−1(U)∩Vx
- So (f|Vx)−1(U)=f−1(U)∩Vx is an open neighbourhood to x∈f−1(U)
- (f|Vx)−1(U) is relatively open in Vx (as we must consider Vx with the subspace topology it inherits from (X,J))
- Since x∈f−1(U) was arbitrary, we see this is true for all x∈f−1(U)
- We notice that for every point, x∈f−1(U), there is an open neighbourhood to x contained in f−1(U).
- Recall a set is open if and only if every point in the set has an open neighbourhood contained within the set
- For any/all x∈f−1(U) there exists an open neighbourhood Vx∈J of x such that f|Vx:Vx→Y is continuous, specifically this means:
TODO: Check this. Also there may be an alternative union construction we could use. Union of the form ⋃x∈f−1(U)(f|Vx)−1(U)
See also
Notes
- Jump up ↑ This denotes the restriction of f to U, so f|U:U→Y by f|U:u↦f(u)
- Jump up ↑ Do I want ⟹ rather than ∧?
TODO: Explore
- Jump up ↑ This is easy to see even without the theorem. If say U is open in Vx then there exists an open set, W, open in X such that U=Vx∩W, but both Vx and W are open in X! So by the properties of being a topology, their intersection is open too.
References