Let’s consider another question: suppose f : D → R is continuous, where D is a subset of R. If (xn) is a sequence in D converging to some real L, is it true that (f(xn)) is also convergent? Now if L is in D, then we know that (f(xn)) → (f(L)). But what if it’s not?
- Let f(x) = 1/x, defined on (0, ∞). Take the sequence xn = 1/n which converges to 0. Then the resulting sequence (f(xn)) = (n) diverges.
- Let f(x) = sin(1/x), defined on (0, ∞). Then the sequence xn = 2/((2n-1)π) converges to 0 but the resulting (f(xn)) = (1, -1, 1, -1, … ) diverges.
Let’s first consider the case where the limit L must lie in D. To that end, we have the following result.
Proposition. For an arbitrary subset , the following are equivalent.
- If (xn) is a sequence in D converging to L, then L is in D.
- D contains all its points of accumulations.
- The complement Dc = R-D is an open subset of R.
(1)→(2) : Let x be a point of accumulation of D. Hence for each ε = 1, 1/2, 1/3, …, there exists , such that 0 < |xn–x| < 1/n for each n. Then (xn)→x must be in L.
(2)→(3) : Let . Since a is not in D it’s not a point of accumulation of D. By definition, this means that there exists ε>0 such that the set of x satisfying 0 < |x–a| < ε does not intersect D. Together with the fact that a is not in D, we see that , so Dc is open.
(3)→(1) : Let (xn)→L. If , an open subset of R, there must be an open ball . On the other hand, since (xn)→L, some must lie within N(L, ε), which is a contradiction. ♦
Definition. We say that a subset is a closed subset if D-U is an open subset of D.
Warning. A subset of D can be both open and closed. E.g. the empty set (or the entire set D) is so.
Useful note: in particular, a closed and bounded subset D of R must have a maximum and minimum. Indeed, since D is bounded, let M be its supremum; we need to show that M lies in D. Suppose it doesn’t. For any positive ε, M-ε is not an upper bound of D, so there exists an element , x > M-ε and x < M. This shows that M is a point of accumulation of D, and by the above proposition, it must lie in D, which contradicts our assumption. Thus M is a maximum of D. The case for minimum is similar.
- Consider the set . It is closed in R since the complement is which is a union of open intervals. Clearly D is bounded. Being both closed and bounded, it has a minimum and a maximum (in this case, 1 and 3 respectively).
- Consider the set . It is not closed because its complement is and the point 2 has no open neighbourhood of R which is wholly contained in E. Or, look at it another way, if it were closed, then since it’s obviously bounded it would contain a maximum and a minimum, but 2 is the supremum of D which is not contained in D itself.
- Consider the set . The complement is which is a union of open intervals and is hence open. But it’s not upper-bounded so it doesn’t have a maximum.
From the above proposition, if D is a closed subset of R and f : D → R is continuous, then any convergent sequence in D gets mapped to a convergent sequence.
- Prove that f : D → R is continuous if and only if for each closed subset C of R, the set is closed in D.
- Find a continuous function f : R → R and a closed subset C of R such that f(C) is not closed in R.
- Prove that a union of finitely many closed subsets of D is still closed in D.
- Prove that an intersection of (possibly infinitely many) closed subsets of D is still closed in D.
- Suppose .
- Prove that if C is closed in D, then B := C ∩ E is closed in E.
- Prove that conversely, if B is closed in E, then B = C ∩ E for some closed subset C of D.
Answers [Highlight to read. Compare with the earlier case of open subsets. ]
- If C is closed in R, then D – f-1(C) = f-1(D – C) is open in D iff f-1(C) is closed in D. Then use the fact that f is continuous iff for each open subset U of R, f-1(U) is open in D.
- Let f be defined by f(x)=1/x for x>1 and f(x)=1 for x≤1. The set C=[1, ∞) is closed in R but f(C) = (0, 1] is not closed in R.
- Follows from the fact that the intersection of finitely many open subsets of D is also open in D.
- Follows from the fact that the union of arbitrarily many open subsets of D is also open in D.
- For (A), if C is closed in D, then D–C is open in D and E–B = E – (C ∩ E) = (D–C) ∩ E is open in E by definition. Thus B is closed in E. For (B), since E–B is open in E, we can write E–B = U ∩ E for some open subset U of D. Then B = E – (U ∩ E) = (D–U) ∩ E where D–U is closed in D.
Now let’s consider the case where D is not closed and (xn) approaches an L not in D.
Definition. The function is uniformly continuous (on D) if:
- for any ε>0, there is a δ>0 such that when x, y in D satisfies |x-y|<δ, we have |f(x)-f(y)|<ε.
Once again we note the difference between standard continuity and uniform continuity. In the former case:
- For every y in D and ε>0, there exists δ>0 such that for all x in D satisfying |x–y|<δ, we have |f(x)-f(y)|<ε.
In the latter case:
- For every ε>0, there exists δ>0 such that for all x, y in D satisfying |x–y|<δ, we have |f(x)-f(y)|<ε.
Using logical quantifiers, these can be succinctly expressed as:
where P(x, y, δ, ε) says “|x–y|<δ implies |f(x)-f(y)|<ε”.
Examples of Uniform Continuity
- Let f(x)=x on any subset S of R. We claim that f is uniformly continuous. Indeed, for any ε>0, let δ=ε. Then whenever |x–y| < δ and x, y in S, we easily get |f(x)-f(y)| = |x–y| < δ = ε, which satisfies the definition for uniform continuity.
- Let f(x)=1/x on (0, ∞). We shall prove that f is not uniformly continuous. Negating the definition of uniform continuity, we need to find an ε>0 such that for any δ>0, we can find x, y>0 with |x–y|<δ but |1/x-1/y|≥ε. So let’s take ε=1; for any δ>0, pick x=δ and y = δ/(δ+1) < x. Then |x–y| = x–y < δ but |1/x-1/y|=1.
- On the other hand, f(x)=1/x on [1, 2] is uniformly continuous. Indeed, whenever 1 ≤ x, y ≤ 2, we have |1/x – 1/y| = |x–y|/|xy| ≤ |x–y| since |xy| ≥ 1. Hence, given any ε>0, we can just let δ=ε. Now whenever x, y in [1, 2] satisfy |x–y| < δ, we also have |f(x)-f(y)| ≤ |x–y| < δ = ε.
Examples 2 and 3 tell us that uniform continuity depends heavily on the domain of the function.
The first main result in this article is:
Theorem. If (xn) is a Cauchy sequence in D and f : D → R is a uniformly continuous function, then (f(xn)) is also a Cauchy sequence.
Proof. Let ε>0.
- Since f is uniformly continuous, pick δ>0 such that when x, y in D satisfies |x–y|<δ, we have |f(x)-f(y)|<ε.
- Since (xn) is Cauchy, pick an N such that when m, n > N, we have |xm–xn|<δ.
This implies that when m, n > N, |f(xm)-f(xn)|<ε. ♦
For example, consider f(x)=1/x on (0, ∞). The sequence is Cauchy but the resulting is not a Cauchy sequence. Hence, this tells us f(x) is not uniformly continuous, as we had verified earlier directly in example 2 above.
Finally, let’s combine uniform convergence and uniform continuity.
Theorem. Suppose is a sequence of uniformly continuous functions on D, which converges uniformly to . Then is also uniformly continuous.
Let ε>0. Then by uniformly convergence, there exists N such that whenever n > N, we have . Fix n. Since is uniformly continuous, there exists δ>0 such that whenever satisfies |x–y|<δ, we also have . Hence:
Now just go back and replace ε by ε/3. ♦