Continuity

Continuity

Definition 33.1. Let $f$ be a function from $X \subset \mathbb{R}$ into $\mathbb{R}$. We say that $f$ is continuous at $a$ if either
(i) $a$ is an accumulation point of $X$ and $\lim_{x \to a} f(x) = f(a)$.
(ii) $a$ is not an accumulation point of $X$.
We say that $f$ is continuous on $[a, b]$ if $f$ is continuous at every point of $[ a, b ]$ (at the end-points $a$ and $b$ we demand that $\lim_{x \to a^{+}} f(x) = f(a)$ and $\lim_{x \to b^{-}} f(x) = f(b)$.)

Theorem 33.2

Theorem 33.2. If $f$ and $g$ are continuous at $a$, then each of the following functions is also continuous at $a$:
(i) $|f|$
(ii) $cf$ for each $c \in \mathbb{R}$
(iii) $f + g$
(iv) $f - g$
(v) $f \cdot g$
(vi) $\dfrac{f}{g}$, provided that $g(a) \ne 0$

Proof. 

Theorem 33.3

Theorem 33.3. Let $f$ be a function from $X \subset \mathbb{R}$ into $\mathbb{R}$ and let $a \in X$. Then $f$ is continuous at $a$ $\iff$ for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $|x - a| < \delta$ and $x \in X$, then $|f(x) - f(a)| < \varepsilon$.

Proof. ($\Longrightarrow$) Suppose that $f$ is continuous at $a$. If $a$ is an accumulation point of $X$, then $\lim_{x \to a} f(x) = f(a)$. Then $\forall \varepsilon > 0, \exists \delta > 0$ such that $x \in X \text{ and } 0 < |x-a| < \delta \Longrightarrow |f(x) - f(a)| < \varepsilon$. If $x \in X$ and $|x-a| < \delta$, then either $x=a$ or $0 < |x-a| < \delta$. If $x = a$, then $|f(x) - f(a)| = 0 < \varepsilon$. If $0 < |x-a| < \delta$, then $|f(x) - f(a)| < \varepsilon$ by the assumption. In either case, $|f(x) - f(a)| < \varepsilon.$
If $x$ is not an accumulation point of $X$, then $\exists \delta > 0$ such that $\{ x \in X \, | \, 0 < |x-a| < \delta \} = \emptyset$. If $x \in X$ and $|x-a| < \delta$, then $x=a$. For any $\varepsilon > 0$, $|f(x) - f(a)| = 0 < \varepsilon$. 
($\Longleftarrow$) Suppose that for every $\varepsilon > 0$, there exists $\delta > 0$ such that if $|x - a| < \delta$ and $x \in X$, then $|f(x) - f(a)| < \varepsilon$. If $a$ is not an accumulation point of $X$, then $f$ is continuous at $a$. If $a$ is an accumulation point of $X$, then by the assumption, $\exists \delta > 0$ such that if $x \in X$ and $|x-a| < \delta$, then $|f(x) - f(a)| < \varepsilon$. Therefore, if $x \in X$ and $0 < |x-a| < \delta$, then $|f(x) - f(a)| < \varepsilon$. Thus $\lim_{x \to a} f(x) = f(a)$, so that $f$ is continuous at $a$. $\blacksquare$

Theorem 33.4

Theorem 33.4. If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$, then $f \circ g$ is continuous at $a$.

Proof. By Theorem 33.3, $\forall \varepsilon > 0, \exists \delta_1 > 0$ such that if $|x-g(a)| < \delta_1$ $\Longrightarrow$ $|f(x) - f(g(a))| < \varepsilon$, and $\exists \delta > 0$ such that $|x-a| < \delta$ $\Longrightarrow$ $|g(x) - g(a)| < \delta_1$. Thus if $|x-a| < \delta$, then $|g(x) - g(a)| < \delta_1$, which implys that $|f(g(x)) - f(g(a))| < \varepsilon$. Thus $f \circ g$ is continuous at $a$ by Theorem 33.3. $\blacksquare$