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**Definition 32**.: A function \(f:S\to\mathbb{R}\) is said to be _bounded_ if its range \(f(S)\) is a bounded subset of \(\mathbb{R}\). That is, \(f\) is bounded if there exists a real number \(M\) such that
\[|f(x)|\leq M\quad\text{for all}\quad x\in S.\]
Unfortunately, a continuous function could be unbounded even if its domain is bounded. For example, set \(S=(0,1)\) and \(f(x)=\dfrac{1}{x}\), and then \(f\) is continuous in its domain, but its range \(f(S)=(1,\infty)\) is an unbounded set. If it happens that the domain of a continuous function is both “closed and bounded”–that is to say a “compact” set (recall from the Heine-Borel Theorem that a subset
Figure 1.35: The set O is open, but \(f^{-1}(O)\) is not openof \(\mathbb{R}\) is compact if and only if it is closed and bounded)–then the function is bounded.
**Theorem 41** (Preservation of Compact Sets).: _Let \(K\) be a compact subset of \(\mathbb{R}\), and let \(f:K\to\mathbb{R}\) be continuous. Then \(f(K)\) is compact as well._
Proof
: Let \(\{U_{\alpha}\}\) be an open cover for \(f(K)\). Since \(f\) is continuous, the inverse image of an open set is open; thus each of the sets \(f^{-1}(U_{\alpha})\) is open. However \(K\) is compact, and therefore every open cover has a finite sub cover, i.e., there are finitely many indices \(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\), such that
\[K\subset f^{-1}(U_{\alpha_{1}})\cup f^{-1}(U_{\alpha_{2}})\cup\cdots\cup f^{-1 }(U_{\alpha_{n}}). \tag{1.2} \]\]
Recalling the fact \(f(f^{-1}(E))\subset E\) for every subset \(E\) in \(\mathbb{R}\), using the inclusion in (1.2), we get
\[f(K)\subset U_{\alpha_{1}}\cup U_{\alpha_{2}}\cup\cdots\cup U_{\alpha_{n}}.\qed\]
Note that the inverse image of a compact set under a continuous function need not be compact. For example, if we set \(E=\{0\}\subset\mathbb{R}\) and \(f:\mathbb{R}\to\mathbb{R}\) to be the function \(f(x)=0\) for all \(x\in\mathbb{R}\), then \(f^{-1}(E)=\mathbb{R}\), which is not compact. The above theorem has important consequences. For example, since compact subsets of \(\mathbb{R}\) are closed and bounded (Heine-Borel Theorem), if \(K\) is a compact subset of \(\mathbb{R}\) and \(f:K\to\mathbb{R}\) is continuous, then \(f(K)\) is closed and bounded. Thus \(f\) is bounded. For the other important consequence of Theorem 41, we need the following definition.
**Definition 33**.: We say that \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) has a _maximum_ at the point \(x_{0}\in S\) if \(f(x)\leq f(x_{0})\) for all \(x\in S\). Similarly, \(f\) has a _minimum_ at the point \(x_{1}\in S\) if \(f(x_{1})\leq f(x)\) for all \(x\in S\).
**Corollary 8** (Extreme Value Theorem).: _If \(f\) is continuous on a closed and bounded set \(K\subseteq\mathbb{R}\), then it attains its maximum and minimum on \(K\); in other words, there are \(x_{0},x_{1}\in K\) such that \(f(x_{1})\leq f(x)\leq f(x_{0})\) for all \(x\in K\)._
Proof
: Let \(K\) be a compact subset of \(\mathbb{R}\), and suppose \(f:K\to\mathbb{R}\) is continuous. From Theorem 41 it follows that \(f(K)\) is compact. Thus \(f(K)\) has both maximum and minimum (\(\inf K=\min K\) and \(\max K=\sup K\) ). Let \(y_{0}=\max f(K)\) and \(y_{1}=\min f(K)\), then there exist \(x_{0},x_{1}\in K\) such that \(y_{0}=f(x_{0})\) and \(y_{1}=f(x_{1})\). It follows that \(f(x_{1})\leq f(x)\leq f(x_{0})\) for all \(x\in K\).
**Remark 22**.: To apply the Extreme Value Theorem, it is essential for \(K\) to be both closed and bounded. If \(K\) is not closed, say \(K=(0,1)\), we can find a continuous function \(f:(0,1)\to\mathbb{R}\) such as \(f(x)=x\), where \(f\) does not assume its max or min values. If \(K\) is not bounded, say \(K=[0,\infty)\), then the identity function \(f:[0,\infty)\to\mathbb{R}\) is continuous but does not assume a maximum value. An example of unbounded continuous function \(f(x)=\frac{1}{x}\) is given in Figure 1.36,
**Remark 23**.: Recall that when we say “\(f\) is continuous on \(S\),” we mean it is continuous at each point \(a\) of \(S\). In other words, given \(\epsilon>0\), we can find \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) whenever \(|x-a|<\delta\). Here \(\delta\) may depend on the point \(a\) as well as on \(\epsilon\), in general \(\delta=\delta(\epsilon,a)\). For example, consider \(f(x)=x^{2}\) and take an arbitrary \(a\in\mathbb{R}\), then \(|f(x)-f(a)|=|x^{2}-a^{2}|=|x-a||x+a|\), and it turns out \(\delta=\frac{\epsilon}{2|a|+1}\) does the trick (why? Fill in the details) (Figure 1.37).
In the above figure for \(f(x)=x^{2}\), we see that a larger \(a\) requires smaller \(\delta\), thus \(\delta=\delta(a,\epsilon)\). There is a stronger concept of continuity called _uniform
Figure 1.37: Larger \(a\) requires smaller \(\delta\)
Figure 1.36: Two continuous functions on a noncompact setcontinuity_, basically requiring that given \(\epsilon>0\), a single \(\delta>0\) can be chosen that works simultaneously for all the points \(a\in S\), i.e., \(\delta=\delta(\epsilon)\) only. A function \(f\) defined on a set \(S\subset\mathbb{R}\) is said to be _uniformly continuous_ on \(S\), if for each \(\epsilon>0\) there is a \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\) for all pairs of point \(x,a\in S\) with \(|x-a|<\delta\). Every uniformly continuous function is continuous (pointwise), but the converse is not true. For example, while \(f(x)=3x\) is uniformly continuous on \(\mathbb{R}\), \(f(x)=x^{2}\) is continuous but not uniformly continuous on \(\mathbb{R}\). However, if the domain of the function is a compact set, then we have the following theorem:
**Theorem 42**.: _Suppose \(f:K\to\mathbb{R}\) is continuous on a compact set \(K\); then \(f\) is uniformly continuous on \(K\)._
For the proof of this theorem, we refer the reader to, e.g., [26], p. 194. Going back to the above example of the continuous function \(f(x)=x^{2}\), if we restrict the domain of this function to some compact set, say \(K=[-3,3]\), then \(|x+a|\leq 6\), thus
\[|f(x)-f(a)|=|x^{2}-a^{2}|=|x-a||x+a|\leq 6\delta=\epsilon.\]
Thus, \(f(x)=x^{2}\) is uniformly continuous on \([-3,3]\).
### Continuous Functions and Connected Sets
Now we turn our attention to connected subsets of \(\mathbb{R}\). The following theorem states that the continuous image of a connected set is connected.
**Theorem 43** (Preservation of Connected Sets).: _Let \(A\) be a connected subset of \(\mathbb{R}\) and \(f:A\to\mathbb{R}\) be a continuous function. Then \(f(A)\) is a connected subset of \(\mathbb{R}\)._
Proof
: The idea of the proof is that if we let \(U\) and \(V\) be open sets in \(\mathbb{R}\) and assume that \(U\) and \(V\) disconnect \(f(A)\), then it is not difficult to show that the sets \(f^{-1}(U)\) and \(f^{-1}(V)\) are open sets (since \(f\) is continuous) in \(\mathbb{R}\) that disconnect \(A\). Suppose \(f(A)\) is not connected. By definition, we can write \(f(A)\subset U\cup V\), where \(f(A)\cap U\neq\emptyset\), \(f(A)\cap V\neq\emptyset\), and \(f(A)\cap U\cap V=\emptyset\). Now, \(f^{-1}(U)=U^{*}\cap A\) for some open set \(U^{*}\), and similarly, \(f^{-1}(V)=V^{*}\cap A\) for some open set \(V^{*}\). From the conditions on \(U\) and \(V\), we obtain that \(U^{*}\cap V^{*}\cap A=\emptyset\), \(A\subset U^{*}\cup V^{*}\), \(U^{*}\cap A\neq\emptyset\), and \(V^{*}\cap A\neq\emptyset\). Thus \(A\) is not connected, in contradiction to the hypothesis.
One consequence of this theorem is the fact that continuous real functions assume all intermediate values on an interval (Figure 1.38).
**Corollary 9** (Intermediate Value Theorem).: _Let \(a
Proof
: Since \([a,b]\) is connected, by Theorem 43 we know \(f([a,b])\) is a connected subset of \(\mathbb{R}\). However, the connected subsets of the real line have a particularly simple structure; they are intervals. A subset \(A\) of the real line \(\mathbb{R}\) is connected if and only if it has the following property: If \(x\in A,\ y\in A\) and \(x The Intermediate Value Theorem could be very useful in deciding about the roots of real-valued continuous functions: **Example 45**.: ### Exercises 1. Use the definition of continuity at a point to prove that: * \(f(x)=3x-5\) is continuous at \(x=2\). Figure 1.38: The Intermediate Value Theorem2. \(f(x)=x^{2}\) is continuous at \(x=3\). 2. A function \(f:\mathbb{R}\to\mathbb{R}\) is called _Lipschitz_ with Lipschitz constant \(\alpha>0\) if \[|f(x)-f(y)|\leq\alpha|x-y|\quad\text{for all }x,y\in\mathbb{R}.\] Give two examples of Lipschitz functions. Moreover, prove that every Lipschitz function is continuous. 3. Let \(f:[a,b]\to\mathbb{R}\) be continuous. Prove that \(|f|\) is also continuous on \([a,b]\). Is the converse true? Namely, if \(|f|\) is continuous on \([a,b]\), is \(f\) also continuous on \([a,b]\)? 4. Let \(f(x)=[x]\) be the greatest integer function that is less than or equal to \(x\), and let \(g(x)=x-[x]\). Sketch the graphs of \(f\) and \(g\). Determine the points at which \(f\) and \(g\) are continuous. 5. Let \(I\) be a closed and bounded subset of \(\mathbb{R}\) and \(f:I\to\mathbb{R}\) be continuous. Prove that \(f\) assumes maximum and minimum values on \(I\), i.e., there exist \(x_{1},x_{2}\in I\) such that \[f(x_{1})\leq f(x)\leq f(x_{2})\] for all \(x\in I\). 6. Prove that if \(f:\mathbb{R}\to\mathbb{R}\) is continuous and periodic, then it attains its supremum and infimum. 7. Let \(f:[a,b]\to\mathbb{R}\) be continuous, and show that \(\sup_{x\in[a,b]}|f(x)|\) is finite. 8. If \(f:\mathbb{R}\to\mathbb{R}\) is continuous and \(K\subset\mathbb{R}\) is connected, is \(f^{-1}(K)\) necessarily connected? 9. Let \(X\) be a compact metric space and \(f:X\to X\) is an isometry, i.e., \[d(f(x),f(y))=d(x,y).\] Show that \(f\) is a bijection. 10. Which of the following functions on \(\mathbb{R}\) are uniformly continuous?
* The cubic function \(f(x)=x^{3}-x-1\) has a real root. Since \(f(1)=-1\) and \(f(2)=5\) and \(-1<0<5\), there exists a \(c\) such that \(f(c)=0\).
* Let \(f:[1,3]\to[0,6]\) be a continuous function satisfying \(f(1)=0\) and \(f(3)=6\). Then there is a point \(c\) in \([1,3]\) such that \(f(c)=c\) (fixed point). To see this define the function \(g(x)=f(x)-x\), and since the difference of two continuous function is continuous, \(g\) is a continuous function. Moreover, \(g(1)=f(1)-1=-1\) and \(g(3)=f(3)-3=3\). Hence, by the Intermediate Value Theorem, \(g\) must vanish at some \(c\in[1,3]\). Thus, \(g(c)=f(c)-c=0\), and this \(c\) is a fixed point of \(f\).
3. \(f(x)=1/x\) is continuous at \(x=1/2\).
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