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Continuous function|Lecture-1

Definition: Let (X, T) and (Y, T*) be topological spaces. A function f from X into Y is continuous relative to T and T*, Or, T-T* continuous or simply continuous, if and only if the inverse image f-1[H] of every T* open subset H of Y is a T – open subset of X, that is, if and only if

                      H T* implies f-1[H] T.

Example: Consider the following topologies on

X = {a, b, c, d} and Y = {x, y, z, w} respectively:

T = {X, ∅, {a}, {a, b}, {a, b, c}}, T* = {Y, ∅, {x}, {y}, {x, y}, {y, z,w}}

Also consider the function f:XY and g: X Y defined by the diagrams

Screenshot_11 Screenshot_12

The function f is continuous since the inverse of each member of the topology T* on Y is a member of the topology T on X. The function g is not continuous since

{y, z, w} T*, i.e., is an open subset of Y, but its inverse image

  g -1[{y, z, w}] = {c, d} is not an open subset of X, i.e., does not belong to T.

Theorem: Let the function f: X Y and g: XY be continuous. Then the composition function g o f : X Z is also continuous.

Proof: Let G be an open subset of Z. Then g -1[G] is open in Y since g is continuous. But f is also continuous, so f -1[g -1[G]] is open in X. Now

[g o f] -1[G] = f -1[g -1[G]]

Thus f -1[g -1[G]] is open in X for every open subset G of Z, or, g o f is continuous.                           (Proved)

Theorem: A function f: XY is continuous if and only if the inverse image of every closed subset of Y is a closed subset of X.

Proof: Suppose f: XY is continuous, and let F be a closed subset of Y. Then Fc is open, and so f -1[Fc] is open in X. But f -1[Fc] = (f -1[F])c; therefore f -1[F] is closed.

Conversely, assume F closed in Y implies f -1[F] closed in X. Let G be an open subset of Y. Then Gc is closed in Y, and so f -1[Gc] = (f -1[F])c is closed in X. Accordingly, f -1[F] is open and therefore f is continuous.  (Proved)

Theorem: Let X and Y be topological spaces. Then a function f:XY is continuous if and only if it is continuous at every point p X.

Proof: Assume f is continuous, and let H Y be an open set containing f(p). But then p f –1 [H], and f –1 [H] is open. Hence f is continuous at p.

Now suppose f is continuous at every point p X, and let H Y be open. For every p f – 1 [H], there exists an open set GpX such that p Gp f –1 [H]. Hence f 1 [H] = ⋃ {Gp: p∊ f –1 [H]} a union of open sets. Accordingly, f –1 [H] is open and so f is continuous. (Proved)

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