COMPACT SETS, CONNECTED SETS AND CONTINUOUS FUNCTIONS
Ali Scagnelli
Fall 2017
Properties of Continuous Functions
The definition below states the most basic yet very important properties of continuous functions. If we have to continuous functions and there is a constant c that is in R, and both f and g are continuous at x0 then the multiplication of c times f or g, addition of f + g, and fg are all continuous at x0 as well.
Let f,g : A → R and let c be an element of R. Suppose f and g are continuous at x0 which is an element of A. Then cf, f + g and fg are continuous at x0. Furthermore, if g(x0) does not equal 0, then f/g is continuous at x0.
Another important definition to know and understand is that of a uniformly continuous function. A function f is uniformly continuous on I when we are able to select δ independently of x0.
Let f be defined on a set A ⊂ R. We say that f is uniformly continuous (on A) if for every ε > 0 there exists δ > 0 such that if x,y ∈ A and |x−y| < δ, then |f(x) − f(y)| < ε.
When the interval I is (closed interval) [a,b] then every function f that is continuous on I is uniformly continuous on I.
Let f be continuous on [a, b]. Then f is uniformly continuous.
We can then use this theorem to prove that any continuous function on a closed bounded interval [a,b] is bounded.
Let f be continuous on [a, b]. Then f possesses both an absolute maximum and an absolute minimum. For us to be able to find and absolute maximum and absolute minimum, f must be continuous on a closed and bounded set. This is supported by the example below where we try to substitute a closed interval with an open interval, and a bounded closed interval with an unbounded closed interval
f(x)=1/x ∈ (0,1)
f(x)= x for x ∈ [0,∞)
https://www.math.ucdavis.edu/~hunter/m125a/intro_analysis_ch3.pdf
This online resource was very helpful with understanding the topic of continuous functions on a compact domain. It comes from the University of California, Davis intro to analysis course. Much of its notation is similar to our textbook. It provided definitions and in depth examples that help better understand this topic of distinguishing whether or not functions are continuous.
Problem on Uniform Continuity see exercise 15.9 http://faculty.atu.edu/mfinan/3203/sol15.pdf
Darboux Property states that if the graph has no point on some horizontal line y = c, then the graph must be entirely above or below that line. In logical terms it is expressed as Theorem 5.53: Let f be continuous on [a,b] and let c ∈ R. If for every x ∈ [a,b], f(x) ̸= c, then either f(x)>c for all x∈[a,b] or f(x)<c for all x∈[a,b].
1. Definitions
1.1. D ⊂ R is compact if and only if for any given open covering of D we can subtract a finite sucovering. That is, given (Gα)α ∈ A a collection of open subsets of R (A an arbitrary set of indices) such that D ⊂∪α∈AGα, then there exists finitely many indices α1, . . . , αN ∈ A such that D ⊂ ∪N i=1Gα1 .
1.2. Let D an arbitrary subset of R. Then A ⊂ D is open in D (or relative to D, or D-open) if and only if there exists G open subset of R such that D = G ∩ D. Similarly we can define the notion of D-closed sets. Note that D is both open and closed in D, and so is ∅.
1.3. D ⊂ R is connected if and only if ∅ and D are the only subsets of D which are both open in D and closed in D. In other words, if D = A ∪ B and A, B are disjoint D-open subsets of D, then either A = ∅ or B = ∅.
1.4. Let D ⊆ R, a ∈ D a fixed element and f : D → R an arbitrary function. By definition, f is continuous at a if and only if the following property holds
∀ ² > 0, ∃ δa(²) > 0 such that |x − a| < δa(²) ∧ x ∈ D ⇒ |f(x) − f(a)| < ²
The last implication can be re-written in terms of sets as follows:
f ³ Ba ¡ δa(²) ¢ ∩ D ´ ⊆ Bf(a)(²)
Here we use the notation Bx(r) := (x − r, x + r).
1.5. Sequence characterization of continuity: f is continuous at a iff for any sequence (xn)n≥1 such that xn ∈ D, n ≥ 1 and limn→∞ xn = a, we have limn→∞ f(xn) = f(a).
1.6. We say that f : D → R is continuous on D (or simply say continuous) if and only if f is continuous at every a ∈ D.
2. Characterization of continuous functions using preimages
2.1. Theorem. Let D ⊆ R and f : D → R a function. Then the following propositions are equivalent:
a) f is continuous (on D).
b) ∀ G ⊆ R open, f −1 (G) is open in D.
c) ∀ F ⊆ R closed, f −1 (F) is closed in D.
Proof. a ⇒ b. Let G ⊆ R open. Pick a ∈ f −1 (G). Then f(a) ∈ G, and since G is open, there must exist ² > 0 such that Bf(a)(²) ⊆ G. By continuity, corresponding to this ² > 0 there exists δ > 0 such that f ³ Ba(δ) ∩ D ´ ⊂ Bf(a)(²). But this places the entire set Ba(δ) ∩ D inside f −1 (G):
Ba(δ) ∩ D ⊆ f −1 (G)
Writing now δ = δa to mark the dependence of δ on a, and varying a ∈ f −1 (G), we obtain
f −1 (G) = ¡ ∪a∈f−1(G)Ba(δa) ¢ ∩ D
which shows that f −1 (G) is open in D.
b ⇔ c. Let F ⊆ R a closed set, which is equivalent to saying that G = CF (the complement in R) is open. Then
f −1 (F) = {x ∈ D|f(x) ∈ F} = {x ∈ D|f(x) ∈/ G} = D − f −1 (G)
Since the complement of a D-open subset of D is D-closed, it means that f −1 (F) is closed in D if and only if f −1 (G) is open in D.
c ⇒ a: exercise.
2.2. Using this characterization, we can prove for example that the composition of continuous functions is a continuous function.
Proposition. Assume f : D → R is continuous, g : E → R is continuous, and f(D) ⊆ E. Then the function h := g ◦ f : D → R defined by h(x) = g(f(x)) is continuous.
Proof. Let G ⊆ R an open set. Then h −1 (G) = f −1 ¡ g −1 (G) ¢ . But g −1 (G) = V ∩ E, for some open set V ⊆ R. But then h −1 (G) = f −1 (V ∩ E) = f −1 (V ) is open in D, so h is continuous.
2.2.1. Example. Assume f : D → R is a continuous function, such that f(x) 6= 0, ∀x ∈ D. Then h : D → R given by h(x) = 1/f(x), is continuous as well. Proof: g : R − {0} → R, g(x) = 1/x is continuous (proved in class), f(D) ⊆ R − {0}, hence h = g ◦ f is continuous.
3. General properties continuous functions
3.1. Theorem. A continuous function maps compact sets into compact sets.
Proof. In other words, assume f : D → R is continuous and D is compact. Then we need to prove that the image f(D) is a compact subset of R. For that, we consider an arbitrary open covering f(D) ⊆ ∪αGα of f(D) and we will try to find a finite subcovering. Taking the preimage we have D ⊆ ∪αf −1 (Gα). But f −1 (Gα) is open in D, so there must exist Vα ⊆ R open such that f −1 (Gα) = Vα ∩ D. Then D ⊆ ∪α(Vα ∩ D) which simply means that D ⊆ ∪αVα. We thus arrived at an open covering of D, so there must exist finitely many indices α1, . . . , αN such that D ⊆ ∪N i=1Vαi , which implies the equality D = ∪ N i=1(Vαi ∩D) = ∪ N i=1f −1 (Gαi ). But this implies in turn that f(D) ⊆ ∪N i=1Gαi . So f(D) is compact.
3.2. Theorem. A continuous function maps connected sets into connected sets.
In other words, assume f : D → R is continuous and D is connected. Then f(D) is connected as well.
Proof. Assume f(D) is not connected. Then there must exist A, B disjoint, non-empty, subsets of f(D), both open relative to f(D), such that f(D) = A ∪ B. Being open relative to f(D) simply means there exists U, V ⊆ R open such that A = f(D) ∩ U, B = f(D) ∩ V . So f(D) ⊆ U ∪ V . But this implies that D ⊂ f −1 (U) ∩ f −1 (V ). Since U, V are open, it follows that f −1 (U) and f −1 (V ) are open relative to D. But they are also disjoint (why?). Since D is connected, it follows that at least one of them, say f −1 (U), is empty. But A ⊆ U, so this forces f −1 (A) = ∅ as well, which is impossible unless A = ∅ (note that A is a subset of the image of f), contradiction.
3.3. Theorem. A continuous function on a compact set is uniformly continuous.
Proof. Assume D compact and f : D → R continuous. Given ² > 0 we need to find δ(²) > 0 such that if x, y ∈ D and |x − y| < δ(²), then |f(x) − f(y)| < ².
From the definition of continuity, given such ² > 0 and x ∈ D, there exists δx(²) such that if |y−x| < δx(²), then |f(y) − f(x)| < ². Clearly D ⊆ ∪x∈DBx( 1 2 δ(²/2)). From this open covering we can extract a finite subcovering (D is compact!), meaning there must exists finitely many x1, x2, . . . , xN ∈ D such that D ⊆ ∪N i=1Bxi ( 1 2 δxi (²/2)).
Let now δ(²) = min{ 1 2 δx1 (²/2), . . . , 1 2 δxN (²/2)}. We will show that δ(²) does the job.
Take y, z ∈ D arbitrary such that |y − z| < δ(²). The idea is that y will be near some xj , which in turn places z near that same xj . But that forces both f(y), f(z) to be close to f(xj ) (by continuity at xj ), and hence close to each other.
Since y ∈ D, there must exist some j, 1 ≤ j ≤ N such that y ∈ Bxj ( 1 2 δxj (²/2)). Thus
• |y − xj | < 1 2 δxj (²/2)
• but |y − z| < δ(²) ≤ 1 2 δxj (²/2)
By the triangle inequality it follows that |z − x| < δxj (²/2). So y, z are within δxj (²/2) of x. This implies that
• |f(y) − f(xj )| < ²/2
• |f(z) − f(xj )| < ²/2
By the triangle inequality once again we have |f(y) − f(z)| < ².
Alternative proof, using sequences. Assume f is not uniformly continuous, meaning that there exists ² > 0 such that no δ > 0 does the job. Checking what this means for δ = 1 n , we see that for any such n ≥ 1 there exist xn, yn ∈ D such that |xn −yn| < 1 n and yet |f(xn)−f(yn)| > ². However D is compact, in particular any sequence in D has a convergent subsequence whose limit belongs to D. Applying this principle twice we find that there must exist n1 < n2 < . . . such that the subsequences (xnk )k≥1 and (ynk )k≥1 are convergent, and x = limk→∞ xnk ∈ D, y = limk→∞ ynk ∈ D. We have the following:
• By construction, |xnk − ynk | < 1 nk ≤ 1 k . Taking the limit, we find x = y.
• By continuity, limk→∞ f(xnk ) = f(x), since x ∈ D. Also limk→∞ f(ynk ) = f(y).
• Also by construction, |f(xnk ) − f(ynk )| > ² hence in the limit, |f(x) − f(y)| ≥ ².
We thus reach a contradiction.
4. Characterization of the compact sets and the connected sets of R
4.1. Theorem. In R, compact = bounded & closed. We prove more, namely:
4.2. Proposition. Let D ⊆ R. Then the following propositions are equivalent:
a) D is compact
b) D is bounded and closed
c) Every sequence in D has a convergent subsequence whose limit belongs to D.
Proof. a ⇒ b. D ⊆ R = ∪n=1∞(−n, n) is an open covering of D. Hence ∃N ≥ 1 such that D ⊆ ∪ N n=1(−n, n) = (−N, N). This shows D is bounded. To prove D is closed, we prove that R − D is open. Let y ∈ R − D. Then D ⊆ ∪∞ n=1(R − [y − 1 n , 1 n ]) (why?). This open covering must have a finite subcovering, so ∃N ≥ 1 such that D ⊆ R−[y − 1 N , y + 1 N ]. But this implies that (y − 1 N , y + 1 N ) ⊆ R− D. But y was chosen arbitrary in R − D, so this set is open, and hence D itself is closed. b ⇒ c. This has to do with the fact that every bounded sequence has a convergent subsequence. c ⇒ b. Here one shows D = D and this has to do with the fact that D is the set of limits of convergent sequences of D, etc… c ⇒ a Let D ⊆ ∪∞ k=1Gk be an arbitrary open covering of D. (Note: a covering by a countable collection of open sets is not the most general infinite open covering one can imagine, of course; we need an intermediate step to prove that from any open covering of D we can extract a countable subcovering, and this has to do with the fact that R admits a countable dense set. Read the details of this step in the textbook.) Let’s prove there exists n ≥ 1 such that D ⊆ ∪∞ n=1Gk. Assume this was not the case. Then ∀n ≥ 1, there exists xn ∈ D − ∪n k=1Gk. But xn is a sequence in D, so it must have a convergent subsequence, call it (xnj )j≥1, with limit in D, so limj→∞ xnj = a ∈ D. But a belongs to one of the G0 i s, say a ∈ GN . Since GN is open, it follows that xnj ∈ GN , for j ≥ j0 (j large enough). In particular this shows that for j large enough (larger than j0 and larger than N) we have xnj ∈ GN ⊆ ∪nj k=1, since nj ≥ j > N. This contradicts the defining property of xn’s.
4.3. Theorem. R is connected.
Proof. This is really the statement that ∅ and R itself are the only subsets of R which are both open and closed. To prove this, let E be a non-empty subset of R with this property. We’ll prove that E = R. For that, take an arbitrary c ∈ R. To prove that c ∈ E, we assume that c /∈ E and look for a contradiction. Since E is non-empty, it follows that E either has points to the left of c or to the right of c. Assume the former holds.
α) Consider the set S = {x ∈ E|x < c}. By construction, S is bounded from above (c is an upper bound for S). Therefore we can consider y = l.u.b.(S) ∈ R.
β) Input: E is closed. Then S = E ∩ (−∞, c] is also closed. Then y ∈ S¯ = S, so y < c.
γ) Input: E is open. y ∈ S ⊆ E and E is open, this means that there exists ² > 0 such that (y−², y+²) ⊆ E. Choose ² small enough so that ² < c − y. In that case z = y + ² 2 ∈ (y − ², y + ²) ⊆ E is an element of E which the properties
• z < c, hence z ∈ S
• z > y
which is in contradiction with the defining property of y.
4.4. Theorem. The only connected subsets of R are the intervals (bounded or unbounded, open or closed or neither).
Proof. First we prove that a connected subset of R must be an interval.
Step 1. Let E ⊆ R a connected subset. We prove that if a ∈ b ∈ E, then [a, b] ⊆ E. In other words, together with any two elements, E contains the entire interval between them. To see this, let c a real number between a and b. Assume c /∈ E. Then E = A∪B, where A = (−∞, c)∩E and B = (c, +∞)∩E. Note that A and B are disjoint subsets of D, both open relative to D. Since E is connected, at least one of them should be empty, contradiction, since a ∈ A and b ∈ B. Thus c ∈ E.
Step 2. To show that E is actually an interval, consider inf E and sup E. Case one. E is bounded. Then m = inf E, M = sup E ∈ R, and clearly E ⊆ [m, M]. On the other hand, for any given x ∈ (m, M), there exists a, b ∈ E such that a < x < b. That’s because m, M ∈ E and one can find elements of E as close to m (resp. M) as desired (draw a picture with the interval (m, M) and place a point x inside it). But then [a, b] ⊆ E, and in particular x ∈ E. Since x was chosen arbitrarily in (m, M), we must have (m, M) ⊆ E ⊆ [m, M], so E is definitely an interval. Case two: E is unbounded. With a similar argument, show that E is an unbounded interval.
Conversely, we need to show that intervals are indeed connected sets. The proof is almost identical to that in the case where the interval is R itself.
5. Corollaries: theorems for continuous functions on R
5.1. Theorem. Let D ⊆ R compact and f : D → R a continuous function. Then there exists y1, y2 ∈ D such that f(y1) ≤ f(x) ≤ f(y2), ∀x ∈ D.
Proof. f(D) is a compact subset of R, so it is bounded and closed. This implies that g.l.b(f(D)) ∈ f(D) and l.u.b.(f(D)) ∈ f(D) as well. But then there must exist y1, y2 ∈ D such that f(y1) = g.l.b.f(D) and f(y2) = l.u.b.f(D). But this implies f(D) ⊆ [f(y1), f(y2)] and we are done.
Note: one uses the notation supx∈D f(x) to denote the l.u.b. of the image of D. In other words, supx∈D f(x) = l.u.b.{f(y)| y ∈ D}. The theorem says that if D is compact and f is continuous, then supx∈D f(x) is finite, and more over that there exists y1 ∈ D such that f(y1) = supx∈D f(y). If the domain is not compact, one can find examples of continuous functions such that either i) sup f = +∞ or such that ii) sup f is a real number but not in the image of f.
For case i), take f(x) = 1/x defined on (0, 1]. For case ii), take f(x) = x defined on [0, 1).
5.2. Theorem. A continuous (real-valued) function defined on an interval in R has the intermediate value property.
Proof. Assume E is an interval in R and f : E → R a continuous function. Let a, b ∈ E (say a < b) and y a number between f(a) and f(b). The intermediate value property is the statement that there exists c between a and b such that f(c) = y. But this follows immediately from the fact that f(E) is an interval. (E is an interval in R ⇒ E is connected ⇒ f(E) is a connected subset of R ⇒ f(E) is an interval in R).
5.3. Problem. Prove that there does not exist a continuous, bijective function f : [0, 1) → R.
Answer. Assume such a function exists. Then f([0, 1 2 ]) is a compact subset of R, so it is bounded. There must exist N > 0 such that f([0, 1 2 ]) ⊆ [−N, N]. But f is assumed to be surjective, so there must exist a, b ∈ [0, 1) such that f(a) = −N − 1 and f(b) = N + 1. Certainly a, b ∈ ( 1 2 , 1). By the intermediate value property [−N − 1, N + 1] ⊆ f(( 1 2 , 1)). In particular f(0) ∈ f(( 1 2 , 1)), which means that there exists c ∈ ( 1 2 , 1) such that f(0) = f(c). But this means f is not injective, contradiction.