Fxto Y is Continuous if and Only if Given Any Closed Subset a in Y F 1 a is Closed

Let's consider another question: supposef :D →R is continuous, whereD is a subset ofR. If (x_n ) is a sequence inD converging to some realL, is it true that (f(x_n )) is also convergent? Now ifL is inD, then we know that (f(x_n )) → (f(L)). But what if it's not?

Examples

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

Closed Subsets

Let's first consider the case where the limitL must lie inD. To that end, we have the following result.

Proposition. For an arbitrary subset $D\subseteq \mathbf{R}$ , the following are equivalent.

If (x_n) is a sequence in D converging to L, then L is in D.

D contains all its points of accumulations.

The complement D^c = R-D is an open subset ofR.

Proof.

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

(2)→(3) : Let $a\in D^c$ . Sincea is not inD it's not a point of accumulation ofD. By definition, this means that there exists ε>0 such that the set ofx satisfying 0 < |x–a| < ε does not intersectD. Together with the fact thata is not inD, we see that $N(a,\epsilon)\subseteq D^c$ , soD^c is open.

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

Definition. We say that a subset $U\subset D$ is aclosed 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, sinceD is bounded, letM be its supremum; we need to show thatM lies inD. Suppose it doesn't. For any positive ε,M-ε is not an upper bound ofD, so there exists an element $x\in D$ ,x >M-ε andx <M. This shows thatM is a point of accumulation ofD, and by the above proposition, it must lie inD, which contradicts our assumption. ThusM is a maximum ofD. The case for minimum is similar.

Examples

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. ClearlyD is bounded. Being both closed and bounded, it has a minimum and a maximum (in this case, 1 and 3 respectively).
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 inE. 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 ofD which is not contained inD itself.
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, ifD is a closed subset ofR andf :D →R is continuous, then any convergent sequence inD gets mapped to a convergent sequence.

Important Exercises

Prove thatf :D →R is continuous if and only if for each closed subsetC of R, the set $f^{-1}(C)$ is closed inD.
Find a continuous functionf :R →R and a closed subsetC ofR such thatf(C) is not closed inR.
Prove that a union of finitely many closed subsets ofD is still closed inD.
Prove that an intersection of (possibly infinitely many) closed subsets ofD is still closed inD.
Suppose .
1. Prove that ifC is closed inD, thenB :=C ∩E is closed inE.
2. Prove that conversely, ifB is closed inE, thenB =C ∩E for some closed subsetC ofD.

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

IfC is closed inR, then D –f ^-1(C) =f ^-1(D –C) is open in D ifff ^-1(C) is closed in D. Then use the fact thatf is continuous iff for each open subsetU ofR,f ^-1(U) is open inD.
Letf be defined byf(x)=1/x forx>1 andf(x)=1 forx≤1. The set C=[1, ∞) is closed inRbutf(C) = (0, 1] is not closed inR.
Follows from the fact that the intersection of finitely many open subsets ofD is also open inD.
Follows from the fact that the union of arbitrarily many open subsets ofD is also open inD.
For (A), ifC is closed inD, thenD–C is open inD andE–B =E – (C ∩E) = (D–C) ∩E is open inE by definition. ThusB is closed inE. For (B), sinceE–B is open inE, we can writeE–B =U ∩E for some open subsetU ofD. Then B =E – (U ∩E) = (D–U) ∩E whereD–U is closed inD.

Uniform Continuity

Now let's consider the case whereD is not closed and (x_n ) approaches anL not inD.

Definition. The function $f:D\to\mathbf{R}$ isuniformly 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 everyy inD and ε>0, there exists δ>0 such that for allx inD 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 allx, y inD satisfying |x–y|<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 $\forall \epsilon,\exists\delta, \forall y, \forall x, P(x,y,\delta,\epsilon)$ ,

whereP(x,y, δ, ε) says "|x–y|<δ implies |f(x)-f(y)|<ε".

Examples of Uniform Continuity

Letf(x)=x on any subset S ofR. We claim thatf is uniformly continuous. Indeed, for any ε>0, let δ=ε. Then whenever |x–y| < δ andx,y inS, we easily get |f(x)-f(y)| = |x–y| < δ = ε, which satisfies the definition for uniform continuity.
Letf(x)=1/x on (0, ∞). We shall prove thatf 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 findx,y>0 with |x–y|<δ but |1/x-1/y|≥ε. So let's take ε=1; for any δ>0, pickx=δ andy= 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 wheneverx,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.

Main Theorems

The first main result in this article is:

Theorem. If (x_n) is a Cauchy sequence in D and f : D →R is a uniformly continuous function, then (f(x_n)) is also a Cauchy sequence.

Proof. Let ε>0.

Sincef is uniformly continuous, pick δ>0 such that whenx,y inD satisfies |x–y|<δ, we have |f(x)-f(y)|<ε.
Since (x_n ) is Cauchy, pick anN such that whenm,n >N, we have |x_m –x_n |<δ.

This implies that whenm,n >N, |f(x_m )-f(x_n )|<ε. ♦

For example, considerf(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 usf(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 existsN such that whenever n >N, we have $||f_n-f||<\epsilon$ . Fixn. Since $f_n:D\to\mathbf{R}$ is uniformly continuous, there exists δ>0 such that whenever $x,y\in D$ satisfies |x–y|<δ, we also have $|f_n(x)-f_n(y)|<\epsilon$ . Hence: