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Textbook by: Lawrence Baggett. E-mail the author

# Continuity and Topology

Module by: Lawrence Baggett. E-mail the author

Summary: This module contains a theorem about the connection between continuity and topology, involving open, closed sets and sequential continuity.

Let f:STf:ST be a function, and let AA be a subset of the codomain T.T. Recall that f-1(A)f-1(A) denotes the subset of the domain SS consisting of all those xSxS for which f(x)A.f(x)A.

Our original definition of continuity was in terms of ϵϵ's and δδ's. (Reference) established an equivalent form of continuity, often called “sequential continuity,” that involves convergence of sequences. The next result shows a connection between continuity and topology, i.e., open and closed sets.

## Theorem 1

1. Suppose SS is a closed subset of CC and that f:SCf:SC is a complex-valued function on S.S. Then ff is continuous on SS if and only if f-1(A)f-1(A) is a closed set whenever AA is a closed subset of C.C. That is, ff is continuous on a closed set SS if and only if the inverse image of every closed set is closed.
2. Suppose UU is an open subset of CC and that f:UCf:UC is a complex-valued function on U.U. Then ff is continuous on UU if and only if f-1(A)f-1(A) is an open set whenever AA is an open subset of C.C. That is, ff is continuous on an open set UU if and only if the inverse image of every open set is open.

### Proof

Suppose first that ff is continuous on a closed set SS and that AA is a closed subset of C.C. We wish to show that f-1(A)f-1(A) is closed. Thus, let {xn}{xn} be a sequence of points in f-1(A)f-1(A) that converges to a point c.c. Because SS is a closed set, we know that cS,cS, but in order to see that f-1(A)f-1(A) is closed, we need to show that cf-1(A).cf-1(A). That is, we need to show that f(c)A.f(c)A. Now, f(xn)Af(xn)A for every n,n, and, because ff is continuous at c,c, we have by (Reference) that f(c)=limf(xn).f(c)=limf(xn). Hence, f(c)f(c) is a limit point of A,A, and so f(c)Af(c)A because AA is a closed set. Therefore, cf-1(A),cf-1(A), and f-1(A)f-1(A) is closed.

Conversely, still supposing that SS is a closed set, suppose ff is not continuous on S,S, and let cc be a point of SS at which ff fails to be continuous. Then, there exists an ϵ>0ϵ>0 and a sequence {xn}{xn} of elements of SS such that c=limxnc=limxn but such that |f(c)-f(xn)|ϵ|f(c)-f(xn)|ϵ for all n.n. (Why? See the proof of Theorem 3.4.) Let AA be the complement of the open disk Bϵ(f(c)).Bϵ(f(c)). Then AA is a closed subset of C.C. We have that f(xn)Af(xn)A for all n,n, but f(c)f(c) is not in A.A. So, xnf-1(A)xnf-1(A) for all n,n, but c=limxnc=limxn is not in f-1(A).f-1(A). Hence, f-1(A)f-1(A) does not contain all of its limit points, and so f-1(A)f-1(A) is not closed. Hence, if ff is not continuous on S,S, then there exists a closed set AA such that f-1(A)f-1(A) is not closed. This completes the proof of the second half of part (1).

Next, suppose UU is an open set, and assume that ff is continuous on U.U. Let AA be an open set in C,C, and let cc be an element of f-1(A).f-1(A). In order to prove thatf-1(A)f-1(A) is open, we need to show that cc belongs to the interior of f-1(A).f-1(A). Now, f(c)A,f(c)A,AA is open, and so there exists an ϵ>0ϵ>0 such that the entire disk Bϵ(f(c))A.Bϵ(f(c))A. Then, because ff is continuous at the point c,c, there exists a δ>0δ>0 such that if |x-c|<δ|x-c|<δ then |f(x)-f(c)|<ϵ.|f(x)-f(c)|<ϵ. In other words, if xBδ(c),xBδ(c), then f(x)Bϵ(f(c))A.f(x)Bϵ(f(c))A. This means that Bδ(c)Bδ(c) is contained in f-1(A),f-1(A), and hence cc belongs to the interior of f-1(A).f-1(A). Hence, if ff is continuous on an open set U,U, then f-1(A)f-1(A) is open whenever AA is open. This proves half of part (2).

Finally, still assuming that UU is open, suppose f-1(A)f-1(A) is open whenever AA is open, let cc be a point of S,S, and let us prove that ff is continuous at c.c. Thus, let ϵ>0ϵ>0 be given, and let AA be the open set A=Bϵ(f(c)).A=Bϵ(f(c)). Then, by our assumption, f-1(A)f-1(A) is an open set. Also, cc belongs to this open set f-1(A),f-1(A), and hence cc belongs to the interior of f-1(A).f-1(A). Therefore, there exists a δ>0δ>0 such that the entire disk bδ(c)f-1(A).bδ(c)f-1(A). But this means that if SS satisfies |x-c|<δ,|x-c|<δ, then xBδ(c)f-1(A),xBδ(c)f-1(A), and so f(x)A=Bϵ(f(c)).f(x)A=Bϵ(f(c)). Therefore, if |x-c|<δ,|x-c|<δ, then |f(x)-f(c)|<ϵ,|f(x)-f(c)|<ϵ, which proves that ff is continuous at c,c, and the theorem is completely proved.

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