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Mathematics LibreTexts

8.5: Continuous Functions

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Continuity

Let (X,dX) and (Y,dY) be metric spaces and cX. Then f:XY is continuous at c if for every ϵ>0 there is a δ>0 such that whenever xX and dX(x,c)<δ, then dY(f(x),f(c))<ϵ.

When f:XY is continuous at all cX, then we simply say that f is a continuous function.

The definition agrees with the definition from when f is a real-valued function on the real line, when we take the standard metric on R.

[prop:contiscont] Let (X,dX) and (Y,dY) be metric spaces and cX. Then f:XY is continuous at c if and only if for every sequence {xn} in X converging to c, the sequence {f(xn)} converges to f(c).

Suppose that f is continuous at c. Let {xn} be a sequence in X converging to c. Given ϵ>0, there is a δ>0 such that d(x,c)<δ implies d(f(x),f(c))<ϵ. So take M such that for all nM, we have d(xn,c)<δ, then d(f(xn),f(c))<ϵ. Hence {f(xn)} converges to f(c).

On the other hand suppose that f is not continuous at c. Then there exists an ϵ>0, such that for every \nicefrac1n there exists an xnX, d(xn,c)<\nicefrac1n such that d(f(xn),f(c))ϵ. Therefore {f(xn)} does not converge to f(c).

Compactness and continuity

Continuous maps do not map closed sets to closed sets. For example, f:(0,1)R defined by f(x):=x takes the set (0,1), which is closed in (0,1), to the set (0,1), which is not closed in R. On the other hand continuous maps do preserve compact sets.

Let (X,dX) and (Y,dY) be metric spaces, and f:XY is a continuous function. If KX is a compact set, then f(K) is a compact set.

Let {f(xn)}n=1 be a sequence in f(K), then {xn}n=1 is a sequence in K. The set K is compact and therefore has a subsequence {xni}i=1 that converges to some xK. By continuity, limif(xni)=f(x)f(K). Therefore every sequence in f(K) has a subsequence convergent to a point in f(K), so f(K) is compact by .

As before, f:XR achieves an absolute minimum at cX if f(x)f(c) for all xX. On the other hand, f achieves an absolute maximum at cX if f(x)f(c) for all xX.

Let (X,d) and be a compact metric space, and f:XR is a continuous function. Then f(X) is compact and in fact f achieves an absolute minimum and an absolute maximum on X.

As X is compact and f is continuous, we have that f(X)R is compact. Hence f(X) is closed and bounded. In particular, supf(X)f(X) and inff(X)f(X). That is because both the sup and inf can be achieved by sequences in f(X) and f(X) is closed. Therefore there is some xX such that f(x)=supf(X) and some yX such that f(y)=inff(X).

Continuity and topology

Let us see how to define continuity just in the terms of topology, that is, the open sets. We have already seen that topology determines which sequences converge, and so it is no wonder that the topology also determines continuity of functions.

[lemma:mstopocontloc] Let (X,dX) and (Y,dY) be metric spaces. A function f:XY is continuous at cX if and only if for every open neighbourhood U of f(c) in Y, the set f1(U) contains an open neighbourhood of c in X.

Suppose that f is continuous at c. Let U be an open neighbourhood of f(c) in Y, then BY(f(c),ϵ)U for some ϵ>0. As f is continuous, then there exists a δ>0 such that whenever x is such that dX(x,c)<δ, then dY(f(x),f(c))<ϵ. In other words, BX(c,δ)f1(BY(f(c),ϵ)). and BX(c,δ) is an open neighbourhood of c.

For the other direction, let ϵ>0 be given. If f1(BY(f(c),ϵ)) contains an open neighbourhood, it contains a ball, that is there is some δ>0 such that BX(c,δ)f1(BY(f(c),ϵ)). That means precisely that if dX(x,c)<δ then dY(f(x),f(c))<ϵ and so f is continuous at c.

[thm:mstopocont] Let (X,dX) and (Y,dY) be metric spaces. A function f:XY is continuous if and only if for every open UY, f1(U) is open in X.

The proof follows from and is left as an exercise.

Exercises

Consider NR with the standard metric. Let (X,d) be a metric space and f:XN a continuous function. a) Prove that if X is connected, then f is constant (the range of f is a single value). b) Find an example where X is disconnected and f is not constant.

Let f:R2R be defined by f(0,0):=0, and f(x,y):=xyx2+y2 if (x,y)(0,0). a) Show that for any fixed x, the function that takes y to f(x,y) is continuous. Similarly for any fixed y, the function that takes x to f(x,y) is continuous. b) Show that f is not continuous.

Suppose that f:XY is continuous for metric spaces (X,dX) and (Y,dY). Let AX. a) Show that f(¯A)¯f(A). b) Show that the subset can be proper.

Prove . Hint: Use .

[exercise:msconnconn] Suppose that f:XY is continuous for metric spaces (X,dX) and (Y,dY). Show that if X is connected, then f(X) is connected.

Prove the following version of the intermediate value theorem. Let (X,d) be a connected metric space and f:XR a continuous function. Suppose that there exist x0,x1X and yR such that f(x0)<y<f(x1). Then prove that there exists a zX such that f(z)=y. Hint: see .

A continuous function f:XY for metric spaces (X,dX) and (Y,dY) is said to be proper if for every compact set KY, the set f1(K) is compact. Suppose that a continuous f:(0,1)(0,1) is proper and {xn} is a sequence in (0,1) that converges to 0. Show that {f(xn)} has no subsequence that converges in (0,1).

Let (X,dX) and (Y,dY) be metric space and f:XY be a one to one and onto continuous function. Suppose that X is compact. Prove that the inverse f1:YX is continuous.

Take the metric space of continuous functions C([0,1]). Let k:[0,1]×[0,1]R be a continuous function. Given fC([0,1]) define φf(x):=10k(x,y)f(y) dy. a) Show that T(f):=φf defines a function T:C([0,1])C([0,1]). b) Show that T is continuous.

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This page titled 8.5: Continuous Functions is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Jiří Lebl via source content that was edited to the style and standards of the LibreTexts platform.

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