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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Note: this might be called the local criterion for continuity

Statement

Let (X,J) and (Y,K) be topological spaces, and let f:XY be a map between them, then[1]:

  • f:XY is continuous if and only if for each point xX there exists an open neighbourhood of x, say U, such that f|U:UY[Note 1] is continuous
    • Symbolically, the right hand side can be written:
      • xXUJ[xU(f|U:UY is continuous)][Note 2]
    • And the full statement

TODO: Symbolical representation of entire statement, including continuity!


Proof

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Continuous Statement

Suppose f:XY is continuous. We wish to show xXUJ[xU(f|U:UY is continuous)]

  • Let xX be given.
    • Then take XJ as our open neighbourhood of x. Automatically xX and f|X:XY is just f:XY and is continuous, we're done.

Statement continuous

Suppose xXUJ[xU(f|U:UY is continuous)], we wish to show f:XY is continuous.

  • Let UK be given (so U is open in (Y,K)), we must show f1(U) is open in (X,J).

TODO: Check this. Also there may be an alternative union construction we could use. Union of the form xf1(U)(f|Vx)1(U)


See also

Notes

  1. Jump up This denotes the restriction of f to U, so f|U:UY by f|U:uf(u)
  2. Jump up Do I want rather than ?

    TODO: Explore


  3. 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=VxW, but both Vx and W are open in X! So by the properties of being a topology, their intersection is open too.

References

  1. Jump up Introduction to Topological Manifolds - John M. Lee