Proof. Clearly, if \(\,\displaystyle {\lim _{x\rightarrow p} f(x) = q}\,\) and \(f\) is continuous at \(p\), then for every \(\varepsilon > 0\, \exists \, \delta > 0 \, \ni \, \begin {vmatrix} f(x) - q\\ \end {vmatrix} = \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). Hence \(\,\displaystyle {\lim _{x\rightarrow p} f(x) = f(p)}\).
Conversely, if \(\, \displaystyle {\lim _{x\rightarrow p} f(x) = f(p)\,}\), then for any given \(\, \varepsilon > 0 \, \exists \, \delta > 0\, \ni \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \, \hspace {0.2cm} \forall \, x \in E\, \) for which \(\, \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). Hence \(f\) is continuous at \(p\in E\).

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Proof. Let \(f\) be continuous at \(p\). Then for every \(\, \varepsilon > 0\, \exists \, \delta > 0 \, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \) whenever \(\, \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \hspace {0.3cm} \cdots \cdots \hspace {0.3cm} (I)\)
Also let \(\displaystyle {\lim _{n\rightarrow \infty } x_n = p}\). Then for \(\delta > 0\, \exists \, N\in \mathbb {N}\, \ni \, \begin {vmatrix} x_n - p\\ \end {vmatrix} < \delta \, \, \forall \, n > \mathbb { N}\hspace {0.3cm} \cdots \cdots \hspace {0.3cm} (II)\).
Putting \(x= x_n\) in \((I)\) we get \(\begin {vmatrix} f(x_n) - f(p)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x_n - p\\ \end {vmatrix} < \delta \).
\(\, \implies \, \begin {vmatrix} f(x_n) - f(p)\\ \end {vmatrix} < \varepsilon \hspace {0.2cm} \forall \, n \> N\hspace {0.3cm}\) (by \((II)\)) \(\, \implies \, \displaystyle {\lim _{n\rightarrow \infty } f(x_n) = f(p)}\)
Conversely, let \(f\) not be continuous at \(p\). Then \(\, \exists \,\) at least one \(\varepsilon > 0\, \ni \, \forall \, \delta > 0\, \exists \,\) a \(\, p\in E\, \ni \, \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \) but \(\begin {vmatrix} f(x) - f(p)\\ \end {vmatrix}\geq \varepsilon \). We take \(\delta = \dfrac {1}{n}\, ,\, n\in \mathbb {N}\). Then for each \(n\, \exists \) and \(x_n\, \begin {vmatrix} x_n - p\\ \end {vmatrix} < \dfrac {1}{n}\hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (III)\) and \(\begin {vmatrix} f(x_n) - f(p)\\ \end {vmatrix}\geq \varepsilon \hspace {0.3cm} \cdots \cdots \hspace {0.3cm} (IV)\). Now, \((III)\) gives \(\displaystyle {\lim _{n\rightarrow \infty } x_n = p}\). But \((IV)\) suggest that \(\displaystyle {\lim _{n\rightarrow \infty } f(x_n) \neq f(p)}\). This is a contradiction. Hence \(f\) is continuous.

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Proof. Note that \(\, \Big \{\dfrac {1}{n}\Big \}^{\infty }_{n=1}\,\) is such that \(\, \displaystyle {\lim _{n\rightarrow \infty }\dfrac {1}{n} = 0}\), but \(\, \displaystyle {\lim _{n\rightarrow \infty } f\Big (\dfrac {1}{n}\Big ) = \lim _{n\rightarrow \infty } = n = \infty }\,\). But \(f(0)\) is not defined. Therefore, by the above corollary \(f\) is discontinuous at \(x = 0\).

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Proof. Each part follows from the corresponding limit law, since continuity at \(p\) says \(\lim _{x\rightarrow p}f(x) = f(p)\).

For the sum, given \(\varepsilon >0\) choose \(\delta _1\) and \(\delta _2\) so that \(\left |f(x)-f(p)\right |<\varepsilon /2\) and \(\left |g(x)-g(p)\right |<\varepsilon /2\) on the respective neighbourhoods; then \(\delta =\min (\delta _1,\delta _2)\) gives \[\left |\left (f+g\right )(x)-\left (f+g\right )(p)\right | \leq \left |f(x)-f(p)\right | + \left |g(x)-g(p)\right | < \varepsilon .\] Scalar multiples are the case \(g\) constant. For the product, write \[fg(x)-fg(p) = f(x)\left [g(x)-g(p)\right ] + g(p)\left [f(x)-f(p)\right ],\] and note \(f\) is bounded near \(p\) because it is continuous there; both terms are then made small. For the quotient the same device applies once \(g(p)\neq 0\) guarantees \(\left |g(x)\right |>\left |g(p)\right |/2\) near \(p\), which keeps the denominator away from zero. □

Proof. Let \(f\) be continuous at \(p\in E\). Then for each \(\, \varepsilon > 0\, \exists \, \delta > 0\, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). Now, \(\begin {vmatrix} \begin {vmatrix} f(x)\\ \end {vmatrix} - \begin {vmatrix} f(p)\\ \end {vmatrix} \end {vmatrix} \leq \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \) whenever \(\begin {vmatrix} x-p\\ \end {vmatrix} < \delta \). Therefore \(\begin {vmatrix} f\\ \end {vmatrix}\) is continuous.

Note that the converse of the above theorem is not generally true. For example, the function \(f\) defined by \[ f(x) = \begin {cases} 1 & x \hspace {0.2cm}\text {is rational}\\ -1 & x \hspace {0.2cm}\text {is irrational}\\ \end {cases}\] is discontinuous everywhere but \(\begin {vmatrix} f\\ \end {vmatrix}= 1\), being a constant function is continuous everywhere.

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Proof. Let \(\varepsilon > 0\) be given. Since \(g\) is continuous at \(f(p),\, \exists \, \eta > 0\, \ni \, \begin {vmatrix} g(y) - g\big (f(p)\big )\\ \end {vmatrix} < \varepsilon \) if \(\begin {vmatrix} y - f(p)\\ \end {vmatrix} < \eta \,\) where \(y \in f(E)\). Since \(f\) is continuous at \(p, \, \exists \, \delta > 0 \, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \eta \,\) if \(\, \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \, , \, x\in E\). Hence, \(\begin {vmatrix} h(x) - h(p)\\ \end {vmatrix} = \begin {vmatrix} g\big (f(x)\big ) - g\big (f(p)\big )\\ \end {vmatrix} < \varepsilon ,\,\) whenever \(\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \) and \(x\in E\). Therefore, \(h\) is continuous at \(p\).

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Proof. Suppose \(f\) is continuous on \(X\) and \(O\) is an open set in \(Y\). We need to show that every point of \(f^{-1}(O)\) is an interior point of \(f^{-1}(O)\). So, suppose \(p\in X\) and \(f(p)\in O\). Since \(O\) is open, \(\, \exists \,\varepsilon \, > 0\, \ni \, y \in O\) if \(\begin {vmatrix} f(x) - y\\ \end {vmatrix} < \varepsilon \), and since \(f\) is continuous at \(p,\, \exists \, \delta \, > 0 \, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \,\) if \(\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). Thus \(\, x \in f^{-1}(O)\,\) as soon as \(\, \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \).
Conversely, suppose \(f^{-1}(O)\) is open in \(X\) for every open set \(O\) in \(Y\). Fix \(p\in X\) and let \(\varepsilon > 0\) be given. Also let \(O\) be the set of all \(y \in Y \,\ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \). Then \(O\) is open. Hence \(\, \exists \, \delta > 0\, \ni \, x\in f^{-1}(O)\) as soon as \(\begin {vmatrix} x - p\\ \end {vmatrix}< \delta \). But if \(x \in f^{-1}(O)\), then \(f(x) \in O\) so that \(\begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \). Thus, for every \(\varepsilon > 0\, \exists \, \delta > 0\, \ni \, \begin {vmatrix} f(x) - f(p)\\ \end {vmatrix} < \varepsilon \,\) whenever \(\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \). Therefore, \(f\) is continuous at \(p\). Hence, \(p\) being arbitrary in \(X\), \(\, f\) is continuous on \(X\).

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Proof. This follows from the theorem sine a set is closed if and only if its complement is open, and \(f^{-1}\big (C^c\big ) = \big [f^{-1}\big (C\big )\big ]^c\) for every \(C\) in \(Y\).


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