testing for continuity via closure operation

Proposition 1.

Let X,Y be topological spacesMathworldPlanetmath, and f:XY a function. Then the following are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

  1. 1.

    f is continuousMathworldPlanetmathPlanetmath,

  2. 2.

    for any closed setPlanetmathPlanetmath DY, the set f-1(D) is closed in X,

  3. 3.

    f(A¯)f(A)¯, where A¯ is the closureMathworldPlanetmathPlanetmath of A,

  4. 4.


  5. 5.

    f-1(C)f-1(C), where C is the interior of C.

  • (1)(2). Use the identity f-1(A-B)=f-1(A)-f-1(B) for any function f. Then f-1(Y-D)=X-f-1(D). So if D is closed (or open), f-1(Y-D) is open (or closed), whence f-1(D) is closed (or open).

  • (2)(3). Suppose first that f:XY is continuous. Since

    f(A)¯={CC closed in Y, and f(A)C},

    Af-1f(A)f-1(C), which is closed in X. So A¯f-1(C), and therefore f(A¯)ff-1(C)C. As a result,

    f(A¯){CC closed in Y, and f(A)C}=f(A)¯.

    Conversely, let V be closed in Y. Then V¯=V. Let U=f-1(V). So f(U)=V. Let W=U¯. Then f(W)=f(U¯)f(U)¯=V¯=V. So Wf-1f(W)f-1(V)=UU¯=W. As a result, U=W is closed.

  • (3)(4). First, assume (2). Let BY and A=f-1(B). So f(A)B. Then f(A¯)f(A)¯B¯. As a result, f-1(B)¯=A¯f-1f(A¯)f-1(B¯).

    Conversely, assume (3). Let AX and B=f(A). So Af-1(B). Then

  • (4)(5). First, assume (3). We use the identity: C=Y-Y-C¯. Then

    f-1(C) = f-1(Y-Y-C¯)=f-1(Y)-f-1(Y-C¯)X-f-1(Y-C)¯
    = X-f-1(Y)-f-1(C)¯=X-X-f-1(C)¯=f-1(C).

    Conversely, assume (4). We use the identity B¯=Y-(Y-B). Then

    f-1(B)¯ = X-(X-f-1(B))=X-f-1(Y-B)

Title testing for continuity via closure operation
Canonical name TestingForContinuityViaClosureOperation
Date of creation 2013-03-22 19:09:11
Last modified on 2013-03-22 19:09:11
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 8
Author CWoo (3771)
Entry type Result
Classification msc 26A15
Classification msc 54C05