## Basic Analysis: Closed Subsets and Uniform Continuity

Let’s consider another question: suppose fD → 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?

Examples

1. Let f(x) = 1/x, defined on (0, ∞). Take the sequence x= 1/n which converges to 0. Then the resulting sequence (f(xn)) = (n) diverges.
2. Let f(x) = sin(1/x), defined on (0, ∞). Then the sequence x= 2/((2n-1)π) converges to 0 but the resulting (f(xn)) = (1, -1, 1, -1, … ) diverges.

## Closed Subsets

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 $D\subseteq \mathbf{R}$, the following are equivalent.

1. If (xn) is a sequence in D converging to L, then L is in D.
2. D contains all its points of accumulations.
3. The complement Dc = R-D is an open subset of R.

Proof.

(1)→(2) : Let x be a point of accumulation of D. Hence for each ε = 1, 1/2, 1/3, …, there exists $x_n\in D$, such that 0 < |xnx| < 1/n for each n. Then (xn)→x must be in L.

(2)→(3) : Let $a\in D^c$. 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 < |xa| < ε does not intersect D. Together with the fact that a is not in D, we see that $N(a,\epsilon)\subseteq D^c$, so Dc is open.

(3)→(1) : Let (xn)→L. If $L\in D^c$, an open subset of R, there must be an open ball $N(L,\epsilon)\subseteq D^c$. On the other hand, since (xn)→L, some $x_n\in D$ must lie within N(L, ε), which is a contradiction. ♦

Definition. We say that a subset $U\subset D$ 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 $x\in D$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.

Examples

1. Consider the set $D = [1, 2]\cup \{3\}$. It is closed in R since the complement is $(-\infty, 1)\cup (2, 3)\cup (3,\infty)$ 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).
2. Consider the set $D = [1, 2)$. It is not closed because its complement is $E=(-\infty, 1)\cup [2, \infty)$ 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.
3. Consider the set $D = \{0\} \cup [1, \infty)$. The complement is $(-\infty, 0)\cup (0, 1)$ 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 fD → R is continuous, then any convergent sequence in D gets mapped to a convergent sequence.

Important Exercises

1. Prove that fD → R is continuous if and only if for each closed subset C of R, the set $f^{-1}(C)$ is closed in D.
2. Find a continuous function f : R → R and a closed subset C of R such that f(C) is not closed in R.
3. Prove that a union of finitely many closed subsets of D is still closed in D.
4. Prove that an intersection of (possibly infinitely many) closed subsets of D is still closed in D.
5. Suppose $E\subseteq D$.
1. Prove that if C is closed in D, then B := C ∩ E is closed in E.
2. Prove that conversely, if B is closed in E, then BC ∩ E for some closed subset C of D.

Answers [Highlight to read. Compare with the earlier case of open subsets. ]

1. 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 Rf-1(U) is open in D.
2. 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 but f(C) = (0, 1] is not closed in R.
3. Follows from the fact that the intersection of finitely many open subsets of D is also open in D.
4. Follows from the fact that the union of arbitrarily many open subsets of D is also open in D.
5. For (A), if C is closed in D, then DC is open in D and EBE – (C ∩ E) = (DC) ∩ E is open in E by definition. Thus B is closed in E. For (B), since EB is open in E, we can write EBU ∩ E for some open subset U of D. Then BE – (U ∩ E) = (DU) ∩ E where DU is closed in D. ## Uniform Continuity

Now let’s consider the case where D is not closed and (xn) approaches an L not in D.

Definition. The function $f:D\to\mathbf{R}$ 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.

Pictorially: 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 |xy|<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 |xy|<δ, we have |f(x)-f(y)|<ε.

Using logical quantifiers, these can be succinctly expressed as: $\forall \epsilon, \forall y, \exists \delta, \forall x, P(x,y,\delta,\epsilon)$ and $\forall \epsilon,\exists\delta, \forall y, \forall x, P(x,y,\delta,\epsilon)$,

where P(xy, δ, ε) says “|xy|<δ implies |f(x)-f(y)|<ε”.

Examples of Uniform Continuity

1. 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 |xy| < δ and xy in S, we easily get |f(x)-f(y)| = |xy| < δ = ε, which satisfies the definition for uniform continuity.
2. 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 xy>0 with |xy|<δ but |1/x-1/y|≥ε. So let’s take ε=1; for any δ>0, pick x=δ and = δ/(δ+1) < x. Then |xy| = xy < δ but |1/x-1/y|=1.
3. On the other hand, f(x)=1/x on [1, 2] is uniformly continuous. Indeed, whenever 1 ≤ xy ≤ 2, we have |1/x – 1/y| = |xy|/|xy| ≤ |xy| since |xy| ≥ 1. Hence, given any ε>0, we can just let δ=ε. Now whenever xy in [1, 2] satisfy |xy| < δ, we also have |f(x)-f(y)| ≤ |xy| < δ = ε.

Examples 2 and 3 tell us that uniform continuity depends heavily on the domain of the function. ## Main Theorems

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 xy in D satisfies |xy|<δ, we have |f(x)-f(y)|<ε.
• Since (xn) is Cauchy, pick an N such that when mnN, we have |xmxn|<δ.

This implies that when mnN, |f(xm)-f(xn)|<ε. ♦

For example, consider f(x)=1/x on (0, ∞). The sequence $x_n = 1/n$ is Cauchy but the resulting $f(x_n) = n$ 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 $f_n:D \to \mathbf{R}$ is a sequence of uniformly continuous functions on D, which converges uniformly to $f:D\to\mathbf{R}$. Then $f:D\to\mathbf{R}$ is also uniformly continuous.

Proof.

Let ε>0. Then by uniformly convergence, there exists N such that whenever nN, we have $||f_n-f||<\epsilon$. Fix n. Since $f_n:D\to\mathbf{R}$ is uniformly continuous, there exists δ>0 such that whenever $x,y\in D$ satisfies |xy|<δ, we also have $|f_n(x)-f_n(y)|<\epsilon$. Hence: $|f(x)-f(y)| \le |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|<\epsilon+\epsilon+\epsilon = 3\epsilon$.

Now just go back and replace ε by ε/3. ♦

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