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\[\lim_{x\to a}f(x)=L,\]
by which we roughly mean that values of \(f(x)\) get arbitrarily close to the real number \(L\) when \(x\) is approaching \(a\). Moreover the point \(a\) need not even be in the domain of \(f\); thus we do not even think of the case when \(x=a\). The following definition is very “similar” to the definition for the limit of a sequence.
**Definition 29** (\(\epsilon-\delta\) Definition of Continuity).: Let \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) and let \(a\) be a limit point of the domain \(S\). We say \(\lim\limits_{x\to a}f(x)=L\) if for each \(\epsilon>0\) there exists a number \(\delta>0\) such that
\[|f(x)-L|<\epsilon\quad\text{whenever}\quad 0<|x-a|<\delta\ (\text{and}\ x\in S).\]
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This follows from sequential criterion of functional limits and the uniqueness of the limit for sequences (see Theorem 38). One very useful application of the sequential criterion for the limit of a function is to show that a given limit does not exist. This is illustrated in the following example.
**Example 40**.: Consider the function \(f(x)=\sin\left(\frac{1}{x}\right)\) for \(x>0\). We claim that \(\lim_{x\to 0}f(x)\) does not exist. Let \(x_{n}=\frac{2}{n\pi}\) for all \(n\in\mathbb{N}\), then \(\lim_{n\to\infty}x_{n}=0\). However,
\[\sin\left(\frac{n\pi}{2}+2\pi k\right)=\left\{\begin{array}{l l}0&\quad \text{if}\quad n=0\\ 1&\quad\text{if}\quad n=1\\ 0&\quad\text{if}\quad n=2\\ -1&\quad\text{if}\quad n=3.\end{array}\right.\]
Clearly \(\{f(x_{n})\}\) is the sequence \(\{1,0,-1,0,1,0,-1,\dots\}\), which does not converge.
### 11 Continuity
The word _continuous_ in everyday use means “unbroken,” “without a break,” or “without gaps.” In mathematics it describes functions for which small changes in the input result in small changes in the output. But these are vague descriptions; continuity needs to be expressed in terms of rigorous mathematical language. The following definition does exactly this.
**Definition 31**.: A real-valued function \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) is said to be continuous at the point \(a\in S\) if for each \(\epsilon>0\) there exists a number \(\delta>0\) such that
\[|f(x)-f(a)|<\epsilon\quad\text{for all}\quad x\in S\quad\text{with}\quad|x-a|<\delta.\]
Written in slightly different terms, this definition requires
\[f(V_{\delta}(a))\subset V_{\epsilon}(f(a)).\]
That is, \(f\) maps a sufficiently small neighborhood of \(a\) into some neighborhood of \(f(a)\). If \(f\) is continuous at every point in the domain \(S\), then we say \(f\) is _continuous on \(S\)_ (Figure 131).
The definition of continuity resembles the definition of functional limits; however, functional limits require the point \(a\) to be a limit point of \(S\), and this is not assumed here. In the above definition it is required that the point \(a\) be in the domain of \(f\); the value \(f(a)\) is then the value of \(\lim_{x\to a}f(x)\). Note also that the above definition implies any function is continuous at isolated points in its domain. Recall that a point of \(S\) that is not a limit point is called an _isolated_ point if there is \(\delta>0\) such that if \(|x-a|<\delta\) and \(x\in S\), then \(x=a\). Thus whenever
\[|x-a|<\delta\text{ and }x\in S,\text{ we have }|f(x)-f(a)|=0<\epsilon.\]
**Example 41**.: Let \(f:\mathbb{R}\to\mathbb{R}\) be the identity function \(f(x)=x\). Fix a point \(x_{0}\in\mathbb{R}\). Given \(\epsilon>0\), we must find \(\delta>0\) such that \(|f(x)-f(x_{0})|<\epsilon\) whenever \(|x-x_{0}|<\delta\). Choosing \(\epsilon=\delta\) does the job.
**Example 42**.: Let
\[f(x)=\left\{\begin{array}{ll}x\sin\left(\frac{1}{x}\right)&:x\neq 0\\ 0&:x=0.\end{array}\right.\]
We claim that \(f\) is a continuous function at \(0\). Observe that
\[|f(x)-f(0)|=|x\sin\left(\frac{1}{x}\right)|\leq|x|\quad\text{ for all}\,x,\]
and given \(\epsilon>0\), we may set \(\delta=\epsilon\). Then when \(|x-0|<\delta\), we have
\[|f(x)-f(0)|\leq|x|<\delta=\epsilon.\]
Hence \(f\) is continuous at \(0\). The graph of this function looks like (Figure 1.32):
Figure 1.31: Continuity of \(f\) at \(a\)
Actually this function is continuous on all points of \(\mathbb{R}\).
**Example 43**.: In discussing continuity one has to pay attention to the domain of the function. For example, the function \(f:\mathbb{R}\to\mathbb{R}\) defined as
\[f(x)=\left\{\begin{array}{ll}0&:x\notin\mathbb{Q}\\ 1&:x\in\mathbb{Q}\end{array}\right.\]
is not continuous at any point of \(\mathbb{R}\). Note that for an arbitrary \(a\in\mathbb{R}\) every neighborhood of \(a\) contains rational points at which \(f(x)=1\) and also irrational points at which \(f(x)=0\). Thus \(\lim\limits_{x\to a}f(x)\) cannot possibly exist. Thus \(f\) is discontinuous at every \(a\in\mathbb{R}\). However, if we restrict \(f\) to be a function from \(\mathbb{Q}\) to \(\mathbb{Q}\), then \(f(x)=1\) is a constant function and is continuous at every point of \(\mathbb{Q}\). It is quite difficult to graph (see our attempt below in Figure 1.33). This function is sometimes referred to as Dirichlet’s function.
### 1.6 Properties of Continuous Functions
It is easy to prove that if \(f\) and \(g\) are defined on a common set \(S\) and are continuous at a point \(a\in S\), then their sum \(f+g\) and product \(fg\) are continuous at \(a\), and the quotient \(\dfrac{f}{g}\) is continuous whenever \(g(a)\neq 0\). Since \(f(x)=x\) is continuous, this allows us to conclude that every polynomial
Figure 1.33: Dirichlet’s function
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**Corollary 6**.: _Let \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) and let \(a\in S\). Then \(f\) is discontinuous at \(a\) if and only if there exists a sequence \(\{x_{n}\}\) in \(S\) such that \(\{x_{n}\}\) converges to \(a\), but the sequence \(\{f(x_{n})\}\) does not converge to \(f(a)\)._
**Example 44**.: Let
\[f(x)=\left\{\begin{array}{ll}\sin\left(\frac{1}{x}\right)&\mbox{ if }\,x\neq 0 \\ 0&\mbox{ if }\,x=0.\end{array}\right.\]
Let \(\{x_{n}\}=\frac{1}{(n+1/2)\pi}\), then for \(x\neq 0\), \(\{f(x_{n})\}=\{\sin(n+1/2)\pi\}=(-1)^{n}\), which is an oscillating sequence. Its limit does not exist and so \(f\) is a discontinuous at \(0\).
Recall that inverse image (or pre-image) of a set \(A\subset\mathbb{R}\) under the function \(f:S\to\mathbb{R}\) is defined to be the set \(\{x\in S:\ f(x)\in A\}\), and usually it is written as \(f^{-1}(A)\). The inverse image of a set under any mapping always makes sense. Although the notation is similar, inverse functions have nothing to do with inverse images. The inverse of a function might not exist. We have seen that the \(\epsilon-\delta\) definition of continuity can be expressed using neighborhoods as \(f(V_{\delta}(a))\subset V_{\epsilon}(f(a))\). Stated in terms of inverse image, our condition reads (Figure 1.34)
\[V_{\delta}(a)\subset f^{-1}(V_{\epsilon}(f(a)).\]
The following theorem illustrates how continuity is related to the pre-image of open subsets in the range of the function. This characterization of continuity is useful not only for real-valued functions but in a more general setting as well. (See [26], p. 180.)
Figure 1.34: Continuity and the inverse image of a set
**Theorem 40**.: _A function \(f:S\subseteq\mathbb{R}\to\mathbb{R}\) is continuous on \(S\) if and only if for any open set \(O\subset\mathbb{R}\) there exists an open set \(G\subset\mathbb{R}\) such that \(f^{-1}(O)=G\cap S\)._
In case where the domain of \(f\) is all of \(\mathbb{R}\), the previous theorem may be restated as follows:
**Corollary 7**.: _A function \(f:\mathbb{R}\to\mathbb{R}\) is continuous if and only if \(f^{-1}(O)\) is open in \(\mathbb{R}\) whenever \(O\) is open in \(\mathbb{R}\)._
For the proof of the above theorem and its corollary, we refer the reader to [26], p. 180. As shown in the following figure, if \(f\) is not continuous, then the inverse image of an open set is not necessarily open (Figure 1.35).
Note that if \(A\subseteq\mathbb{R}\) is an open subset in the domain of \(f\), even if \(f\) is continuous, we cannot claim that image \(f(A)\) is open. A typical counterexample is the continuous function \(f(x)=x^{2}\). If we set \(A=(-1,1)\), then \(f(A)=[0,1)\), which is not open.
### 1.3 Continuous Functions and Compact Sets
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