Properties of Continuous Functions
A function f \colon D \to \mathbb{R} is called bounded if its range f(D) is a subset of \mathbb{R} where there exists some M \in \mathbb{R} such that |f(x)| \leq M for all x \in D.
A functions domain being bounded has no correlation
If D is a compact subset of \mathbb{R} and f \colon D \to \mathbb{R} is continuous, then f(D) is compact.
Let U = \{U_i\} be an open cover of f(D). Since f is continuous on D, by Theorem 2.2.5, there exists some V_i such that V_i \cap D = f^{-1}(U_i). Since f(D) \subseteq \bigcup U_i, we have D \subseteq \bigcup f^{-1}(U_i) \subseteq \bigcup V_i so \{V_i\} is an open cover of D.
Since D is compact, there exists finitly many V_1,\dots,V_n such that D \subseteq V_1 \cup \cdots \cup V_N hence D \subseteq (V_1 \cap D) \cup \cdots \cup (V_N \cap D) which implies f(D) \subseteq U_1 \cup \cdots \cup U_N, thus \{U_1,\ldots,U_n\} is a finite subcover of U. Hence f(D) is compact.
Let D a compact subset of \mathbb{R} and f \colon D \to \mathbb{R} be continuous. Then, f attains a minimum and maximum on D.
We know from Theorem 2.3.1, the image f(D) is compact. All compact sets have both a minimum, y_1 and maximum y_2. Since y_1,y_2 \in f(D), there exists x_1,x_2\in D such that f(x_1) = y_1 and f(x_2) = y_2. Thus f(x_1) \leq f(x) \leq f(x_2) for all x \in D.
On the real number line, a set is compact if and only if it is closed and bounded. We use the definition of open subcovers as it applies to more generalized spaces.
Let f \colon [a,b] \to \mathbb{R} be continuous, and suppose f(a) < 0 < f(b). Then there exists c \in (a,b) such that f(c) = 0.
Let S = \{x \in [a,b] \colon f(x) \leq 0\}. Notice S is bounded above by b, and by our assumuption a \in S. Since S is not empty, and S is bounded, We know c = \sup S exists in [a,b]. We want to show f(c) = 0.
Case 1: Suppose f(c) < 0.
WTS: This assumption contradicts c being a supremum of S, because c is not an upper bound of S.
Since f is continuous, there exists a neighborhood U of c such that f(x) < 0 for all x \in U \cap [a,b] (Prove). Since we’re assumming f(c) < 0 < f(b), we have that c \not = b, and thus there exists some p \in U such that c<p<b. Since p \in U we have f(p) < 0. This contradicts our assumption that c is an upper bound for S.
Case 2: Suppose f(c) > 0.
WTS: This assumption contradicts c being a supremum of S, because c is not a least upper bound of S.
If f(c) > 0, there exists a neighborhood U of c such that f(x) > 0 for all x \cup [a,b], since f(a) < 0 < f(c). We have c \not = a and thus U contains some point, p. We have a < p < c. Since f(x) > 0 for x \in U, no points of S are in [p,c]. This implies p is an upperbound for S and contradicts c being the least upper bound of s.
We conclude then that f(c) = 0. Since f(a) < 0 < f(b) and f(c) = 0, it must be that c \in (a,b)
Let f \colon [a,b] \to \mathbb{R} be continuous. For any value k between f(a) and f(b), there exists c \in (a,b) such that f(c) = k.
Let k be any real number between f(a) and f(b). If f(a) is less than f(b), we apply the previous lemma to a new function given by g(x) = f(x) - k.
Then g(a) = f(a) - k < 0. Then g(b) = f(b) - k > 0. We can then apply the lemma from before to guarentee the existence of c \in (a,b) such that g(c) = 0. This is equivalent to saying that f(c) = k.
In the reverse case, the same argument works where g(x) = k - f(x).
Let I be a compact interval and suppose f \colon I \to \mathbb{R} is a continuous function. Then the set f(I) is a compact interval
Corollary 2.3.2 implies there exists x_1,x_2 \in I such that f(x_1) < f(x) < f(x_2) for all x \in I. Let f(x_1) = m_1 and f(x_2) = m_2. Then clearly f(I)\subseteq [m_1,m_2].
Consider the case m_1 < m_2, as m_1 = m_2 is trivial. For each k \in [m_1,m_2], the Intermediate Value Theorem implies k \ f(c) for some c between x_1,x_2. Thus (m_1,m_2) \subseteq f(I). Since m_1,m_2 \in f(I), we have [m_1,m_2] \subseteq f(I). Hence f(I) is the compact interval [m_1,m_2].
Review
In this section we proved
If D is a compact subset of \mathbb{R} and f \colon D \to \mathbb{R} is continuous, then f(D) is compact.
Let D a compact subset of \mathbb{R} and f \colon D \to \mathbb{R} be continuous. Then, f attains a minimum and maximum on D.
Let f \colon [a,b] \to \mathbb{R} be continuous, and suppose f(a) < 0 < f(b). Then there exists c \in (a,b) such that f(c) = 0.
Let f \colon [a,b] \to \mathbb{R} be continuous. For any value k between f(a) and f(b), there exists c \in (a,b) such that f(c) = k.
Let I be a compact interval and suppose f \colon I \to \mathbb{R} is a continuous function. Then the set f(I) is a compact interval