Limits of Functions
Let $X$ and $Y$ be metric spaces; suppose $E \subset X,$ $f : E \to Y,$ and $p$ is a limit point of $E.$ We write $f(x) \to q$ as $x \to p,$ or
$$ \lim_{x \to p} f(x) = q $$
if there is a point $q \in Y$ with the following property: For every $\epsilon > 0,$ there exists a $\delta > 0$ such that
$$ d_Y(f(x), q) < \epsilon $$
for all points $x \in E$ for which
$$ 0 < d_X(x,p) < \delta. $$
In the above definition, the symbols $d_X$ and $d_Y$ refer to distances in $X$ and $Y,$ respectively.
Furthermore, while $p \in X,$ $p$ need not be a point of $E.$ In fact, even if $p \in E,$ it's possible that $f(p) \neq \lim_{x \to p} f(x).$
Let $X$ and $Y$ be metric spaces; suppose $E \subset X,$ $f : E \to Y,$ and $p$ is a limit point of $E.$ Then
$$ \lim_{x \to p} f(x) = q \tag{a} $$
if and only if
$$ \lim_{n \to \infty} f(p_n) = q \tag{b} $$
for every sequence $\{p_n\}$ in $E$ such that
$$ p_n \neq p, \quad \lim_{n \to \infty} p_n = p. \tag{c} $$
Suppose that (a) is true and let $\{p_n\} \subset E$ be a sequence such that (c) holds. Let $\epsilon > 0.$ Then, for some $\delta > 0,$ if $d_X(x,p) < \delta,$ then $d_Y(f(x), q) < \epsilon.$ Now, there is also some $N$ such that $d_X(p_n, p) < \delta$ whenever $n \geq N,$ and thus, $d_Y(f(p_n), q) < \epsilon$ whenever $n \geq N,$ so (b) holds.
Conversely, suppose that (a) is false. Then, for some $\epsilon > 0,$ that for every $\delta > 0$ there exists a point $x \in E$ such that $d_Y(f(x), q) \geq \epsilon$ but $0 < d_X(x, p) < \delta.$ If we let $\delta_n = 1/n, n = 1, 2, 3, \dots,$ and each $x_n$ a point such that $0 < d_X(x_n, p) < \delta_n,$ then $\{x_n\}$ is a sequence satisfying (c). However, since $d_Y(f(x_n), q) \geq \epsilon$ for all $n,$ (b) does not hold.
$\square$If $f$ has a limit at $p,$ this limit is unique.
This follows directly from the facts that limits of sequences are unique and that limits of functions are characterized by limits of sequences.
$\square$Referenced by (2 direct)
Suppose $E \subset X,$ a metric space, $p$ is a limit point of $E,$ $f$ and $g$ are complex functions on $E,$ and
$$ \lim_{x \to p} f(x) = A, \quad \lim_{x \to p} g(x) = B. $$
Then:
$$ \lim_{x \to p} (f + g)(x) = A + B. $$
$$ \lim_{x \to p} (fg)(x) = AB. $$
$$ \lim_{x \to p} (\frac{f}{g})(x) = \frac{A}{B}, B \neq 0. $$
This follows directly from the that limits of functions are characterized by limits of sequences, and the algebraic properties of sequences of limits.
$\square$If $\vec{f}$ and $\vec{g}$ map $E$ into $R^k,$ then (a) remains true, and (b) becomes
$$ \lim_{x \to p} (f \cdot g)(x) = A \cdot B. $$
That is, the limit of an inner product of vector valued functions is the inner product of their limits.
(See A sequence in $R^k$ converges iff its components converge).
Continuous Functions
Suppose $X$ and $Y$ are metric spaces, $E \subset X, p \in E,$ and $f : E \to Y.$ Then $f$ is said to be continuous at $p$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$ d_Y(f(x), f(p)) < \epsilon $$
for all points $x \in E$ for which $d_X(x, p) < \delta.$
If $f$ is continuous at every point of $E,$ then $f$ is said to be continuous on $E$.
Referenced by (13 direct, 1 transitive)
Direct references:
- note-11
- proof-of-function-is-continuous-at-point-iff-limit-at-point-equals-function-at-point
- function-is-continuous-at-point-iff-limit-at-point-equals-function-at-point
- proof-of-composition-of-continuous-functions-is-continuous
- composition-of-continuous-functions-is-continuous
- mapping-continuous-iff-inverse-images-of-open-sets-are-open
- smooth
- Continuously Differentiable
- theorem-19
- Homeomorphism
- Volume Integral
- Divergence Theorem of Gauss
- Stoke's Theorem
Transitive (depth 1):
A more geometric way to view continuity is that for any ball $B(f(p))$ centered at $f(p),$ there is some ball $B(p)$ centered at $p$ for which every point in $B(p)$ (including $p$) is mapped by $f$ to some point in $B(f(p)).$
Also, note that if $p$ is an isolated point of $E,$ then the definition of continuous implies that every function $f$ which has $E$ as its domain of definition is continuous at $p.$ This is because for any $\epsilon > 0$ we choose, we can pick $\delta > 0$ so that the only point $x \in E$ for which $d_X(x, p) < \delta$ is $x = p,$ and so
$$ d_Y(f(x), f(p)) = 0 < \epsilon. $$
Also note that, unlike the definition of limit, the definition of continuous requires $f$ to be defined at $p$ in order to be continuous at $p.$
Suppose $X$ and $Y$ are metric spaces, $E \subset X, p \in E,$ with $p$ a limit point of $E,$ and $f : E \to Y.$ Then, $f$ is continuous if and only if $\lim_{x \to p} f(x) = f(p).$
Note that the definition of a function having a limit at point in a metric space is different from the definition of a function being continuous at a point in a metric space only in that the continuous definition requires the function to be defined at the point (and equal to the limit at the point.)
$\square$Suppose $X, Y, Z$ are metric spaces, $E \subset X,$ $f: E \to Y,$ $g: f(E) \to Z,$ $h: E \to Z$ with
$$ h(x) = g(f(x)) \quad (x \in E). $$
The function $h$ is called the composition or the composite of $f$ and $g.$ The notation
$$ h = g \circ f $$
is frequently used.
Suppose $X, Y, Z$ are metric spaces, $E \subset X,$ $f: E \to Y,$ $g: f(E) \to Z,$ $h: E \to Z$ with
$$ h(x) = g(f(x)) \quad (x \in E). $$
If $f$ is continuous at a point $p \in E$ and if $g$ is continuous at the point $f(p),$ then $h$ is continuous at $p.$
Let $\epsilon > 0.$ Since $g$ is continuous at $f(p),$ there exists $\eta > 0$ such that
$$ d_Z(g(y), g(f(p))) < \epsilon \text{ if } d_Y(y, f(p)) < \eta \text{ and } y \in f(E). $$
Since $f$ is continuous at $p,$ there exists $\delta > 0$ such that
$$ d_Y(f(x), f(p)) < \eta \text{ if } d_X(x, p) < \delta \text{ and } x \in E. $$
It follows that
$$ d_Z(h(x), h(p)) = d_z(g(f(x)), g(f(p))) < \epsilon $$
if $d_X(x, p) < \delta$ and $x \in E.$ Thus, $h$ is continuous at $p.$
$\square$Basically, since $g$ is continuous, we can control how close its output is to $g(f(p))$ by controlling how close its input is to $f(p),$ which we can certainly do, since $f$ is also continuous, and we can control how close its output is to $f(p)$ by controlling how close its input is to $p.$
A mapping $f$ of a metric space $X$ into a metric space $Y$ is continuous on X if and only if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$ (see inverse image.)
Assume $f$ is continuous on $X$ and $V$ is an open set in $Y.$ Suppose, for the sake of contradiction, that $f^{-1}(V)$ is not open. Then, some point $p \in f^{-1}(V)$ is not an interior point of $f^{-1}(V),$ which means there is no neighborhood of $p$ that contains only points in $f^{-1}(V),$ that is, every neighborhood of $p$ contains some point $q$ that is not in $f^{-1}(V),$ i.e., $f(q) \notin V.$ Now, since $V$ is open and $f(p) \in V,$ there is some $\epsilon > 0$ for which $B_\epsilon(f(p)) \subset V,$ but, since all neighborhoods of $p$ contain some $q \notin f^{-1}(V),$ there is no $\delta > 0$ for which all points within $\delta$ of $p$ are mapped to $B_\epsilon(f(p))$ by $f,$ a contradiction, since $f$ is continuous on $X$ by hypothesis. Therefore, our assumption is incorrect and $f^{-1}(V)$ is open.
Conversely, suppose $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y.$ Let $V$ be an open set in $Y.$ Assume, for the sake of contradiction that there is some $p \in f^{-1}(V)$ at which $f$ is not continuous. Let $\epsilon > 0.$ Then, there is no $\delta > 0$ for which $B_\delta(p)$ contains only points that are mapped by $f$ to $V,$ i.e. every neighborhood of $p$ contains some point that is not in $f^{-1}(V),$ and therefore $p$ is not an interior point of $f^{-1}(V),$ and $f^{-1}(V)$ is not open, a contradiction. Thus, our assumption must be incorrect, and there is no such $p,$ and since entire space $Y$ is an open subset of itself, $f$ must be continuous on all of $X.$
$\square$