8

FURTHER TOPOLOGICAL RESULTS

8.1THE EXTENSION PROBLEM

In this chapter we consider a variety of questions related to the topology of metric spaces. First, we examine the problem of extending a bounded continuous function from a closed subset to the entire space. The following result essentially solves the problem.

8.1.1 THEOREM. Let A and B be disjoint closed subsets of the metric space Ω. There is a continuous f: Ω → [0, 1] such that f = 0 on A and f = 1 on B.

This result holds for a wider class of topological spaces (“normal spaces”), and the general result is known as Urysohn’s Lemma.

    Proof. If C ⊆ Ω, let d(x, C) be the distance from x to C; that is, d(x, C) = inf_{y ∈ C} d(x, y). As shown in Section 4.4.2, the mapping x → d(x, C) is uniformly continuous on Ω. If x ∈ C, then d(x, C) = 0; if x ∉ C and C is closed, then d(x, C) > 0. Thus, if

f has the desired properties. ■

Note that if g(x) = (b − a) f (x) + a, then g is a continuous mapping of Ω into [a, b], and g = a on A, g = b on B.

Before getting to the extension question, we must raise an issue that we have avoided so far, that of relative topology. If Ω is a metric space and A ⊆ E ⊆ Ω, we know what it means to say that A is an open subset of Ω: if x ∈ A, there is an r > 0 such that {y ∈ Ω; d(x, y) < r} ⊆ A. But A is also a subset of E, and if we regard E as the whole space, we then have a definition of A being open in E: if x ∈ A, there is an r > 0 such that {y ∈ E : d(x, y) < r) ⊆ A. Thus, if A is open in Ω, A must be open in E. The converse is not true, however. For example, A = {(x, y) : y = 0, a < x < b} is open in {(x, y) : y = 0, −∞ < x < ∞} (this just says, essentially, that an open interval is an open subset of R), but not open in R². For if a < x < b and y = 0, no open ball centered at (x, y) is a subset of A.

Similarly, if B ⊆ E ⊆ Ω, B is closed in Ω if x_n ∈ B, x_n → x ∈ Ω implies x ∈ B; B is closed in E if x_n ∈ B, x_n → x ∈ E implies x ∈ B. Thus, if B is closed in Ω, then B is closed in E, but again not conversely. For example, (0, 1] is closed in (0, ∞) but not in R.

Under certain conditions, a set open (or closed) in the smaller space will be open (or closed) in the larger space.

8.1.2 THEOREM

        (a) If A is open in E and E is open in Ω, then A is open in Ω.

        (b) If B is closed in E and E is closed in Ω, then B is closed in Ω.

    Proof

        (a) If x ∈ A, there is an r > 0 such that {y ∈ E : d(x, y) < r} ⊆ A. Since E is open in Ω, there is an s > 0 such that {y ∈ Ω : d(x, y) < s} ⊆E. Thus, {y ∈ Ω : d(x, y) < min(r, s)} ⊆ A ∩ E = A.

        (b) Let x_n ∈ B, x_n → x ∈ Ω. Since B ⊆ E and E is closed, x ∈ E. Since B is closed in E, x ∈ B. ■

We may now attack the extension problem.

8.1.3Tietze Extension Theorem

Let f: E → R, where E is a closed subset of the metric space Ω, and f is bounded and continuous on E. There is a continuous g: Ω → R such that g = f on E and sup{|g(x)| : x ∈ Ω} = sup{|f(x)| : x ∈ E}. Thus, f can be extended to the entire space without increasing the bound.

    Proof. Let M = sup_{x ∈ E} |f(x)| < ∞, and define

By continuity of f, A₁ and B₁ are closed in E, and hence in Ω by Theorem 8.1.2(b). By Theorem 8.1.1 there is a continuous function such that on on B₁.

Now f₂ = f − h₁ is continuous on E. On A₁, and ; on and ; on and . Thus, on E.

Continuing, we define

Just as above, we find a continuous such that on on B₂. If f₃ = f₂ − h₂ = f − h₁ − h₂, then f₃ is continuous on E and .

Inductively, we obtain continuous real-valued functions h_n on Ω such that on Ω and

By the Weierstrass M-test, converges uniformly to a continuous function g on Ω, where and f = g on E. ■

Problems for Section 8.1

    1.If A and F are subsets of E, show that if A is open in E then A ∩ F is open in F.

    2.If A and F are subsets of E and A is closed in E, show that A ∩ F is closed in F.

    3.Let f : E → R, where E is a closed subset of the metric space Ω. (Here we do not assume f bounded, as in Theorem 8.1.3.) Show that there is a continuous g : such that g = f on E.

(One way to proceed is to find a homeomorphism, i.e., a one-to-one, onto, continuous map h with a continuous inverse, mapping to a closed, bounded interval of R, and apply Theorem 8.1.3. Another application of this idea is to make into a metric space via (x, y) = |h(x) − h(y)|.)

8.2BAIRE CATEGORY THEOREM

We now turn to a different question. Suppose we wish to construct a function f : R → R that is continuous at each irrational point and discontinuous at each rational point. This can be done explicitly; if r₁, r₂, … is an enumeration of the rationals, let f(r_n) = a_n, where a_n > 0 and a_n → 0 as n → ∞. If x is irrational, define f(x) = 0. Now if x is an irrational number and ∊ > 0, we have 0 < a_n < ∊ for all sufficiently large n, say for n ≥ N. Choose δ > 0 so small that none of the rational numbers r_n, n < N, belong to (x − δ, x + δ). If y ∈ (x − δ, x + δ), then either y is irrational or y = r_n for some n ≥ N. In any event, 0 ≤ f(y) < ∊. Thus, f is continuous at x. If x is a rational r_k, let x₁, x₂, … be a sequence of irrationals converging to x. Then f(x_n) = 0 for all n, but f(x) = a_k > 0. Therefore, f is discontinuous at x, as desired.

Perhaps surprisingly, it is not possible to find a function f: R → R that is continuous at each rational point and discontinuous on the irrationals. This is one of the consequences of the Baire Category Theorem, which we are about to consider. The theorem is concerned with sets that might be called “thin,” namely sets that are nowhere dense.

8.2.1Definitions and Comments

If A is a subset of the metric space Ω, A is said to be nowhere dense if its closure Ā has empty interior. (In general, the interior of a set B, denoted by B⁰, is defined as the union of all open subsets of B. Thus, for A to be nowhere dense, there can be no open subsets of Ā except . We should point out that for our present purposes, “open” always means “open in Ω.”)

For example, any finite set is nowhere dense. Also, we saw in Theorem 4.4.1(e) that the Cantor set is nowhere dense. Note that the rationals Q do not form a nowhere dense subset of R, since .

It follows from the definition that A is nowhere dense if and only if (Ā)^c is dense. For if A is not nowhere dense, let V be a nonempty open subset of Ā. If x ∈ V, no sequence in (Ā)^c can converge to x, for if so the sequence would eventually be in V and hence in Ā. Thus, (Ā)^c is not dense. On the other hand, if A is nowhere dense, x ∈ Ω, and r > 0, the nonempty open set B_r(x) cannot be contained in Ā. Take r = 1/n to obtain points x_n with d(x, x_n) < 1/n and x_n ∉ Ā. Then x_n → x, proving (Ā)^c dense.

To summarize: A is not nowhere dense iff there is an open ball B_r(x) ⊆ Ā iff there is an x that cannot be approximated arbitrarily closely by points in (Ā)^c iff (Ā)^c is not dense.

The set B ⊆ Ω is said to be of category 1 in Ω if B can be expressed as a countable union of nowhere dense subsets of Ω. Otherwise B is said to be of category 2 in Ω.

Now, the main result.

8.2.2Baire Category Theorem

Let Ω be a nonempty, complete metric space. If where the A_n are closed subsets of Ω, then the interior of A_n is nonempty for some n. Therefore, Ω is of category 2 in itself

    Proof. Assume that the interior of A_n is empty for every n. The key idea is to find a sequence of open balls B_δn (x_n) satisfying 0 < and B_δn (x_n) ⊆ Β_δn−1/2(x_{n − 1})\Α_n (hence ). Suppose we can do this. Then x_n ∈ Β_{δn − 1/2}(x_{n − 1}), so d(x_n, x_{n − 1}) < ; consequently (if n < m),

Thus, {x_n} is a Cauchy sequence. By the completeness hypothesis, x_n converges to a limit x ∈ Ω. Now if k > n,

Thus, x_k ∈ Β_δn/2(x_n) for all k > n; hence, . We conclude that x ∈ Β_δn(x_n) for all n, so for all n, x ∉ A_n. This contradicts the assumption that .

To obtain the required sequence of open balls, note that since , we must have A₁ ≠ Ω (because Ω is a nonempty open subset of itself). Therefore, is a nonempty open set, so there is an open ball with . Since is not a subset of A₂; hence, Β_δ1/2(x₁)\Α₂ is a nonempty open set and, therefore, contains an open ball Β_δ2(x₂) with (see Fig. 8.2.1). Similarly, Β_δ2/2(x₂) is not a subset of A₃, so we obtain Β_δ3(x₃) ⊆ Β_δ2/2(x₂)\Α₃, where . Continuing inductively, we obtain the desired sequence. ■

If U_n is an open dense subset of the metric space Ω, then is closed and is dense; hence, is nowhere dense. If Ω is nonempty and complete, then by Theorem 8.2.2, ; that is, . In fact, a stronger result holds.

8.2.3 THEOREM. If U_n is an open dense subset of the complete metric space Ω(n = 1, 2, … ), then is dense.

    Proof. There is nothing to prove if Ω is empty, so assume Ω is nonempty. Let B be any open ball; the closure is a closed set and therefore is a complete metric space. (If {x_n} is a Cauchy sequence in , then x_n converges to a limit x ∈ Ω and, necessarily, x ∈ because is closed.)

Figure 8.2.1 Proof of the Baire Category Theorem

We claim that U_n ∩ B is dense in . For if x ∈ , we can find y_n ∈ B with d(x, y_n) < l/2n, and then find z_n ∈ U_n ∩ B with d(y_n, z_n) < 1/2n (since U_n is dense in Ω). Thus, d(x, z_n) < 1/n → 0, proving U_n ∩ B dense in . Also, U_n ∩ B is open in Ω and therefore open in B. By the remarks before the statement of the theorem, . But then intersects every open ball, so if x ∈ Ω we can find . Thus, . We conclude that is dense. ■

The following result is another direct consequence of the Baire category theorem.

8.2.4 THEOREM. The set of irrational numbers cannot be expressed as a countable union of closed sets; hence_y the rationals cannot be expressed as a countable intersection of open sets.

    Proof. Let I be the set of irrationals and Q the set of rationals. If with all C_n closed (in R), then each C_n is nowhere dense. To see this, note that any nonempty open subset of C_n must contain an open interval and therefore must contain rational numbers, contradicting C_n ⊆ I. Thus, C_n has empty interior. Hence, I is a countable union of nowhere dense sets, as is Q (if {r_n} is an enumeration of the rationals, . Therefore, R is a countable union of nowhere dense sets, contradicting Theorem 8.2.2. ■

A countable union of closed sets is often referred to as an F_σ set; a countable intersection of open sets is called a G_δ set.

The problem of finding a function continuous on the rationals and discontinuous on the irrationals may now be attacked. The following way of expressing the idea of continuity at a point will be helpful.

8.2.5LEMMA. Let f: Ω → Ω′, where Ω and Ω′ are metric spaces. If x is a point of Ω, then f is continuous at x if and only if for every positive integer n there is a δ > 0 such that if x₁, x₂ ∈ Ω and d(x₁, x) < δ, d(x₂, x) < δ, we have d(f(x₁), f(x₂)) < 1/n.

    Proof. If f is continuous at x, let us choose δ > 0 so that d(y, x) < δ implies d(f(y), f(x)) < 1/2n. If d(x₁, x) < δ and d(x₂, x) < δ, then d(f(x₁), f(x₂)) ≤ d(f(x₁), f(x)) + d(f(x), f(x₂)) < l/n. Conversely, if the given condition holds, take x₂ = x to show continuity at x. ■

The key result is the following.

8.2.6 THEOREM. Let f be an arbitrary mapping from the metric space Ω to the metric space Ω′. Then {x ∈ Ω : f is discontinuous at x} is an F_σ, that is, a countable union of closed sets. Thus, by Theorem 8.2.4, there is no f:R → R that is continuous at each rational point and discontinuous at each irrational point.

    Proof. By Lemma 8.2.5, f is continuous at x if and only if

Thus, f is discontinuous at x if and only if (∃n)(∀δ > 0)(∃x₁, x₂ ∈ Ω) such that

Figure 8.2.2 Proof of Theorem 8.2.6

Therefore, {x : f is discontinuous at x} = where

We are finished if we can show that the D_n are closed. Let x_j ∈ D_n, x_j → x. Given δ > 0, we have x_j ∈ B_δ(x) for all sufficiently large j. Pick any such j, and let N be an open ball centered at x_j and entirely contained in B_δ(x), say N = B_δ′(x_j) (see Fig. 8.2.2). Since x_j ∈ D_n, there are points x_j1, x_j2 ∈ N with d(f(x_j1), f(x_j2)) ≥ 1/n. But then (since N ⊆ B_δ(x), so that d(x_j1, x) and d(x_j2, x) are less than δ) x ∈ D_n, proving D_n closed. ■

Problems for Section 8.2

    1.If B_r(x) is an open ball in a metric space, show that

    2.Give an example of a metric space for which the inclusion in Problem 1 is proper; i.e., verify that {y : d(x, y) ≤ r] can be strictly larger than .

    3.Let f₁, f₂, … be real-valued functions on the metric space Ω. Show that

8.3CONNECTEDNESS

Often in analysis, sets occur that consist of several “parts” or “pieces.” One solves a problem on each piece separately and then puts the results together. A topological question that is of interest is “When is a set in one piece?” For example, the set E = (0, 1) (1, 2) seems to have two parts, and one observation that can be made is that (0, 1) and (1, 2) are each open and closed in E. (Note that (0,1) is not closed in R since, for example, 1 − 1/n → 1 ∉ (0, 1). But 1 ∉ E, and in fact there is no sequence in (0, 1) converging to a point of E outside (0, 1). Thus (0, 1) is closed in E.) This idea is used to define a general notion of connectedness.

8.3.1Definitions

Let E be a subset of the metric space Ω. We say that E is disconnected if E can be expressed as A B, where A and B are disjoint, nonempty, and both open (equivalently, both closed) in E. A set is called connected if it is not disconnected.

A somewhat more natural notion of connectedness is the following.

The set E is path-connected if for all x, y ∈ E, there is a path in E joining x to y ; that is, there is a continuous f : [0, 1] → E such that f (0) = x, f (1) = y. If we think of t as a time variable, we are at the point x at time t = 0, at y at time t = 1, and we move smoothly in E from x to y as time goes by.

Path-connectedness is stronger than connectedness, as we now show.

8.3.2 THEOREM. If E is path-connected, then E is connected. Thus, all intervals of R^p are connected, as is R^p itself. (An interval of R^p is defined in the natural way, for example, if a = (a₁, …, a_p), b = (b₁, …,b_p), then

    Proof. Assume E = A B, where A and B are disjoint, nonempty, and closed in E. Pick x ∈ A, y ∈ B, and let f : [0, 1] → E be continuous, with f(0) = x, f(1) = y. Let S = {t ∈ [0, 1] : f(0 ∈ A}, and let t₀ = sup S; note t₀ > 0 because A is open in E. Since t₀ − 1/n is not an upper bound of S, we can find t_n ∈ S with t₀ − 1/n < t_n < t₀. Thus, t_n → t₀ and f(t_n) ∈ A; since f is continuous and A is closed in E, we have f(t_n) → f (t₀) ∈ A. Now t₀ < 1 because f(1) = y ∈ B, and since t₀ is an upper bound of S, f(t) ∈ B for all t > t₀ (with t ∈ [0, 1]). Consider any sequence of points approaching t₀ from above. Since f is continuous and B is closed in . This contradicts the disjointness of A and B. ■

An example of a set that is connected but not path-connected is the “topologist’s sine curve” E = F G, where F={(x, y) ∈ R : y = sin 1/x,0 < x ≤ 1,}, G = {(0, 0)}.

The following result is often useful in proving connectedness.

8.3.3 THEOREM. Let E be the union of the connected sets F and G, and assume that or . Then E is connected.

    Proof. We may as well assume . Suppose A is open and closed in E. Then A ∩ F is open and closed in F (Section 8.1, Problem 1), so A ∩ F = 0 or F. Similarly, or G.

    Case 1. . Then A = A ∩ E = (A ∩ F) (A ∩ G) = 0.

    Case 2. A ∩ F = F, A ∩ G = G. Then F ⊆ A, G ⊆ A; hence E = F G ⊆ A, so A = E.

    Case 3. A ∩ F = F, A ∩ G = . Let z ∈ ∩ G, and let z_n ∈ F, z_n → z. Then z_n ∈ A, z_n → z ∈ G (hence z ∉ A), contradicting the fact that A is closed in E.

    Case 4. , A ∩ G = G. If A₀ = E\A, then (since G ⊆ A) , and A₀ ∩ F = F. (Note that implies F ⊆ A^c, and, by hypothesis, F ⊆ E. Thus F ⊆ A₀.) The situation is now the same as in Case 3, and a contradiction results.

We conclude that or A = F, proving E connected. ■

Problems for Section 8.3

    1.The subsets A and B of the metric space Ω are said to be separated iff A and B are disjoint and both closed (hence open) in A B.

            Thus, E is disconnected iff E can be expressed as the union of two nonempty separated sets. Show that A and B are separated if and only if

    2.Continuing Problem 1, show that A and B are separated if and only if there exist open sets G₁, G₂, with A ⊆ G₁, B ⊆ G₂, and .

    (The conditions given in Problems 1 and 2 are useful because they express the idea of connectedness without using the relative topology.)

    3.Let E be a connected subset of R, and let a = inf E, b = sup E. If a < x < b, show that x ∈ E. Thus, the only connected subsets of R are the intervals.

    4.If f is a continuous function on the connected set E, show that the image f(E) is connected.

    5.Use Problems 3 and 4 to give an alternative proof of the Intermediate Value Theorem 4.3.2.

8.4SEMICONTINUOUS FUNCTIONS

A standard example of a sequence of continuous functions f_n converging pointwise to a discontinuous limit f is

Although the limit function is discontinuous, it does have a “semicontinuity” property, as follows.

8.4.1Definitions and Comments

Let f: , where Ω is a metric space. We say that f is lower semicontinuous (LSC) on Ω if for every , {x ∈ Ω : f(x) > a} is open; f is upper semicontinuous (USC) on Ω if for every a ∈ R, {x ∈ Ω : f(x) < a} is open.

Possibly a useful way of remembering the definition is to note that the function f given above is LSC on R; it takes the lower value at the discontinuity. Also, in the set {x : f (x) > a} involved in the definition of lower semicontinuity, a is the lower number in the inequality. Similarly, if g (x) = 0, x < 0; g(x) = 1, x ≥ 0, then g is USC on R. It takes the upper value at the discontinuity, and in the set {x : g(x) < a} involved in the definition of upper semicontinuity, a is the upper number in the inequality.

A function f : is continuous on Ω if and only if it is both USC and LSC. To see this, note that if f is USC and LSC, then for all a b ∈ , a < b, the set {x : a < f(x) < b} is open. If V is an arbitrary open subset of , then V is a union of open intervals I_n; hence, is open.

In this argument we have glossed over (except in Section 8.1, Problem 3) the fact that we do not yet have a metric on . (If we use ordinary Euclidean distance, we get d(x, ∞) = ∞ for every x ∈ R, which is awkward.) The easiest way out is to identify with a closed bounded interval of reals, for example, with the interval [0, π] by means of the one-to-one onto mapping h(x) = π/2 + arctan x, x ∈ . If x, y ∈ , define d(x, y) to be |h(x) − h(y)|. Then V will be open in if and only if h(V) is open in [0, π]. In particular, (a, ∞] and [−∞, a) are open in .

Note that f is LSC if and only if −f is USC; this often allows proofs about LSC functions to apply to USC functions as well.

The following criterion for semicontinuity is useful.

8.4.2 THEOREM. Let f: Ω → . Then f is LSC on Ω if and only if for each x ∈ Ω and every sequence {x_n} in Ω with x_n → x, we have lim_n inf f(x_n) ≥ f(x). Similarly, f is USC on Ω if and only if x_n → x implies lim_n sup f(x_n) ≤ f(x).

    Proof. Let f be LSC. If x_n → x and b is any number less than f(x), then V = {y ∈ Ω : f(y) > b} is open, and x ∈ V. For large enough n, x_n ∈ V; hence f(x_n) > b. But this implies that any convergent subsequence of {f(x_n)} has a limit that is at least b. We conclude that lim„ inf f(x_n) ≥ b. But b may be chosen arbitrarily close to f(x), and therefore lim„ inf f(x_n) ≥ f(x).

Conversely, assume that x_n → x implies lim_n inf f(x_n) ≥ f(x), and let W = {x : f(x) > a}. We show that W^c is closed (hence W is open). Let x_n ∈ W^c, x_n → x. Then f(x_n) ≤ a for all n, and lim_n inf f(x_n) ≥ f(x) by assumption. Therefore f(x) ≤ a; that is, x ∈ W^c. Thus W^c is closed.

The USC result may be proved in a similar fashion or by applying the LSC result to −f. ■

The condition of Theorem 8.4.2 may be used, if desired, to define upper and lower semicontinuity at a point x ∈ Ω.

The next result is a direct consequence of the definition of semicontinuity.

8.4.3 THEOREM. If {f_i, i ∈ I} is an arbitrary family of LSC functions, then sup_{i ∈ I} f_i is LSC. If f₁, …, f_n are LSC, then min_1≤i≤n f_i is LSC.

Similarly, the inf of an arbitrary family of USC functions is USC, and the maximum of a finite number of USC functions is USC.

(All of the above functions are defined pointwise; for example, if h = sup_{i ∈ I} f_i, then for each x we have h(x) = sup{f_i(x) : i ∈ I}.)

    Proof. Assume f_i LSC for each i, and let f = sup_i f_i. If a ∈ , then which is open. If g = min_1≤i≤n f_i, then , again an open set. The proof for USC functions is analogous. ■

We have seen that a continuous real-valued function on a compact set has a maximum and a minimum. Part of this result is retained in the semicontinuous case.

8.4.4 THEOREM. If f is LSC on the compact metric space Ω, then f attains a minimum; that is, for some x ∈ Ω we have f(x) = inf_y∈Ω f(y).

Similarly, an USC function on a compact metric space attains a maximum.

    Proof. Assume f LSC. If b = inf_y∈n f(y), there is a sequence of points x_n ∈ Ω such that f(x_n) → b. Since Ω is compact, there is by Theorem 2.3.1 a subsequence x_nk converging to a limit x ∈ Ω. By Theorem 8.4.2, lim_k inf f(x_nk) ≥ f(x), and therefore b ≥ f(x). But f(x) > b by definition of b, so f(x) = b. The USC case is handled similarly. ■

If f has a jump at x and, say, f(x⁻) < f(x) < f(x⁺), then, intuitively, f(x⁻) looks like the lim inf of f(y) as y approaches x. In other words, among all possible limits of sequences f(x_n) as x_n → x, f(x⁻) is the smallest. Similarly, f(x⁺) looks like the lim sup of f(y) as y → x. The formalization of these ideas leads to the discovery of the largest LSC function that is less than or equal to a given function f, and the smallest USC function greater than or equal to f.

8.4.5 THEOREM. Let f : Ω → be arbitrary. If x ∈ Ω, let N (x) denote the class of open sets containing x (also called the class of neighborhoods of x). Define

(Note the analogy with the formulas lim inf_n→∞ x_n = sup_n in_k≥n x_k, lim sup_n→∞ x_n = inf_n sup_k≥n x_k. Intuitively, inf_y∈V1 f(y) ≤ inf_y∈V2 f(y) if V₂ ⊆ V₁, so in computing lim inf_y→n f(y) we can ignore V₁. So in a sense, the V ∈ N (x) are approaching x.)

Define the lower and upper envelopes of f by

Then w LSC on Ω and if g is any LSC function on Ω such that g ≤ f, we have g ≤ . Similarly, is USC on Ω, and if g is any USC function on Ω with g ≥ f, then g ≥ . Thus, is the largest LSC function ≤ f, and f is the smallest USC function ≥ f.

    Proof. As usual we consider only the LSC case; to prove the USC result, use a similar argument or apply the LSC result to −f.

Let x_n → x. We wish to show that lim_n inf ; by Theorem 8.4.2, will be LSC on Ω.

Suppose lim_n inf . If V is any neighborhood of x, then there is an x_n ∈ V with f(x_n) < b. (There is a subsequence with ; since x_nk → x we have x_nk ∈ V for large k. Note that n depends on V.) But V is an open set containing x_n, so

so for every V ∈ N (x), inf_y∈V f(y) < b. Thus,

a contradiction.

Now let g LSC, g ≤ f. We claim that lim inf_y→x g(y) ≥ g(x). For if sup_{V ∈ N (x)} inf_y∈V g(y) < g(x), then for some b < g(x) we have inf_y∈V g(y) < b for all V ∈ N(x). Take V = B_1/n (x), and select x_n ∈ B_1/n (x) with g(x_n) < b. Thus, x_n → x, but lim_n inf g(x_n) ≤ b < g(x), contradicting Theorem 8.4.2. The proof is completed by the observation that

If f₁, f₂, … are continuous, extended real-valued functions on Ω and f₁ < f₂ ≤ …, then f = lim_n→∞ f_n is LSC by Theorem 8.4.3 (note f = sup_n f_n). Similarly, if the f_n decrease to f, then f is USC, since in this case f = inf_n f_n. We now prove that any semicontinuous function can be expressed as a monotone limit of continuous functions. (We adopt the following convenient abbreviations. If f₁ ≤ f₂ ≤ … and f_n → f pointwise on Ω, we write f_n ↑ f. If f₁ ≥ f₂ ≥ … and f_n → f, we write f_n ↓ f.)

8.4.6 THEOREM. Let f be LSC on the metric space Ω. Then there is a sequence of continuous functions f_n: Ω → with f_n ↑ f. Similarly, if f is USC, there is a sequence of continuous f_n with f_n ↓ f. If |f| ≤ M < ∞ on Ω, the f_n can be chosen so that |f_n| ≤ M for all n.

    Proof. Again, we consider only the LSC case. First assume f nonnegative and finite valued. Define

If x, y ∈ Ω, then f(z)+ n d(x, z) ≤ f(z) + n d(y, z) + n d(x, y). Take the inf over z (first on the left side, then on the right side) to obtain fn(x) ≤ f_n(y) + n d(x, y). By symmetry,

hence, f_n is uniformly continuous on Ω. Furthermore, since f ≥ 0, we have 0 ≤ f_n(x) ≤ f(x) + n d(x, x) = f(x). By definition, f_n increases with n; we must show that lim_n f_n is actually f.

Given ∊ > 0, by definition of f_n(x) there is a point z_n ∈ Ω such that

But f_n(x) + ∊ ≤ f(x) + ∊; hence d(x, z_n) → 0. Since f is LSC, Theorem 8.4.2 yields lim_n inf f(z_n) ≥ f(x); hence

By (1) and (2),

for all sufficiently large n. Thus, f_n(x) → f(x).

If |f| ≤ M < ∞, then f + M is LSC, finite-valued, and nonnegative. If 0 ≤ g_n ↑ f + M, then f_n = g_n − M ↑ f and |f_n| ≤ M.

In general, observe that h(x) = π/2 + arctan x, x ∈ , is a one-to-one onto mapping of onto [0, π], continuous, and having a continuous inverse (such a mapping is called a homeomorphism). Also, x < y if and only if h(x) < h(y), so that h is order-preserving. Let f₀(x) = h(f(x)); then {x : f₀(x) > a} = {x : f(x) > /h⁻¹(a)}, so f₀ is LSC. Furthermore, f₀ is finite-valued and nonnegative. By what we have proved above, there is a sequence of continuous functions g_n such that g_n(x) ↑ f₀(x) = h(f(x)), x ∈ Ω. If f_n(x) = h⁻¹(g_n(x)), then f_n(x) ↑ h⁻¹(h(f(x))) = f(x), as desired. ■

Problems for Section 8.4

    1.Give an example of an infinite sequence of LSC functions f_n such that inf_n f_n is not LSC.

    2.Give an example of a (nonmonotone) limit of continuous functions that is neither USC nor LSC.

    3.Give an example of a LSC function on a compact set such that f has no maximum.

    4.Formulate an epsilon-delta statement that is equivalent to the condition of Theorem 8.4.2, and prove your result.

    5.In Theorem 8.4.6, if f is LSC and nonnegative, show that the functions f_n can also be taken to be nonnegative.

REVIEW PROBLEMS FOR CHAPTER 8

    1.Let Ω be a metric space and let A ⊆ E ⊆ Ω. If A is open in Ω, then A is open in E

        (a)always

        (b)sometimes

        (c)never

    2.In the previous problem, if A is open in E then A is open in Ω

        (a)always

        (b)sometimes

        (c)never

    3.Give an example of

        (a)a countably infinite nowhere dense subset of R.

        (b)an uncountably infinite nowhere dense subset of R.