
# 4.1: Basic Definitions

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We shall now consider functions whose domains and ranges are sets in some fixed (but otherwise arbitrary) metric spaces $$(S, \rho)$$ and $$\left(T, \rho^{\prime}\right),$$ respectively. We write

$f : A \rightarrow\left(T, \rho^{\prime}\right)$

for a function $$f$$ with $$D_{f}=A \subseteq(S, \rho)$$ and $$D_{f}^{\prime} \subseteq\left(T, \rho^{\prime}\right) . \quad S$$ is called the domain space, and $$T$$ the range space, of $$f .$$

I. Given such a function, we often have to investigate its "local behavior" near some point $$p \in S .$$ In particular, if $$p \in A=D_{f}(\text { so that } f(p) \text { is defined) we }$$ may ask: Is it possible to make the function values $$f(x)$$ as near as we like ($$" \varepsilon -$$ near") to $$f(p)$$ by keeping $$x$$ sufficiently close $$\left(\text { "close }^{\prime \prime}\right)$$ to $$p,$$ i.e., inside some sufficiently small globe $$G_{p}(\delta) ?$$ If this is the case, we say that $$f$$ is continuous at $$p .$$ More precisely, we formulate the following definition.

Definition

A function $$f : A \rightarrow\left(T, \rho^{\prime}\right),$$ with $$A \subseteq(S, \rho),$$ is said to be continuous at $$p$$ iff $$p \in A$$ and, moreover, for each $$\varepsilon>0$$ (no matter how small) there is $$\delta>0$$ such that $$\rho^{\prime}(f(x), f(p))<\varepsilon$$ for all $$x \in A \cap G_{p}(\delta) .$$ In symbols,

$(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in A \cap G_{p}(\delta)\right)\left\{\begin{array}{l}{\rho^{\prime}(f(x), f(p))<\varepsilon, \text { or }} \\ {f(x) \in G_{f(p)}(\varepsilon)}\end{array}\right.$

If $$(1)$$ fails, we say that $$f$$ is discontinuous at $$p$$ and call $$p$$ a discontinuity point of $$f .$$ This is also the case if $$p \notin A$$ (since $$f(p)$$ is not defined).

If $$(1)$$ holds for each p in a set $$B \subseteq A,$$ we say that $$f$$ is continuous on $$B .$$ If this is the case for $$B=A,$$ we simply say that $$f$$ is continuous.

Sometimes we prefer to keep $$x$$ near $$p$$ but different from $$p .$$ We then replace $$G_{p}(\delta)$$ in $$(1)$$ by the set $$G_{p}(\delta)-\{p\},$$ i.e., the globe without its center, denoted $$G_{\neg p}(\delta)$$ and called the deleted $$\delta$$ -globe about $$p .$$ This is even necessary if $$p \notin D_{f}$$. Replacing $$f(p)$$ in $$(1)$$ by some $$q \in T,$$ we then are led to the following definition.

Definition

Given $$f : A \rightarrow\left(T, \rho^{\prime}\right), A \subseteq(S, \rho), p \in S,$$ and $$q \in T,$$ we say that $$f(x)$$ tends to $$q$$ as $$x$$ tends to $$p(f(x) \rightarrow q \text { as } x \rightarrow p)$$ iff for each $$\varepsilon>0$$ there is $$\delta>0$$ such that $$\rho^{\prime}(f(x), q)<\varepsilon$$ for all $$x \in A \cap G_{\neg p}(\delta) .$$ In symbols,

$(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in A \cap G_{\neg p}(\delta)\right) \quad\left\{\begin{array}{l}{\rho^{\prime}(f(x), q)<\varepsilon, \text { i.e. }} \\ {f(x) \in G_{q}(\varepsilon)}\end{array}\right.$

This means that $$f(x)$$ is $$\varepsilon$$ -close to $$q$$ when $$x$$ is $$\delta$$ -close to $$p$$ and $$x \neq p$$.

If $$(2)$$ holds for some $$q,$$ we call $$q$$ a limit of $$f$$ at $$p .$$ There may be no such $$q$$. We then say that $$f$$ has no limit at $$p,$$ or that this limit does not exist. If there is only one such $$q(\text { for a given } p),$$ we write $$q=\lim _{x \rightarrow p} f(x) .$$

Note 1. Formula (2) holds "vacuously" (see Chapter 1,8 §§1-3, end remark) if $$A \cap G_{\neg p}(\delta)=\emptyset$$ for some $$\delta>0 .$$ Then any $$q \in T$$ is a limit at $$p,$$ so a limit exists but is not unique. (We discard the case where $$T$$ is a singleton.)

Note 2. However, uniqueness is ensured if $$A \cap G_{\neg p}(\delta) \neq \emptyset$$ for all $$\delta>0,$$ as we prove below.

Observe that by Corollary 6 of Chapter 3, §14, the set $$A$$ clusters at $$p$$ iff

$(\forall \delta>0) \quad A \cap G_{\neg p}(\delta) \neq \emptyset . \quad(\text { Explain! })$

Thus we have the following corollary.

corollary $$\PageIndex{1}$$

If $$A$$ clusters at $$p$$ in $$(S, \rho),$$ then a function $$f : A \rightarrow\left(T, p^{\prime}\right)$$ can have at most one limit at $$p ;$$ i.e.

$\lim _{x \rightarrow p} f(x) \text{ is unique (if it exists).}$

In particular, this holds if $$A \supseteq(a, b) \subset E^{1}(a<b)$$ and $$p \in[a, b]$$.

Proof

Suppose $$f$$ has $$t w o$$ limits, $$q$$ and $$r,$$ at $$p .$$ By the Hausdorff property,

$G_{q}(\varepsilon) \cap G_{r}(\varepsilon)=\emptyset \quad \text{ for some } \varepsilon>0.$

Also, by $$(2),$$ there are $$\delta^{\prime}, \delta^{\prime \prime}>0$$ such that

$\begin{array}{ll}{\left(\forall x \in A \cap G_{\neg p}\left(\delta^{\prime}\right)\right)} & {f(x) \in G_{q}(\varepsilon) \text { and }} \\ {\left(\forall x \in A \cap G_{\neg p}\left(\delta^{\prime \prime}\right)\right)} & {f(x) \in G_{r}(\varepsilon)}\end{array}$

Let $$\delta=\min \left(\delta^{\prime}, \delta^{\prime \prime}\right) .$$ Then for $$x \in A \cap G_{\neg p}(\delta), f(x)$$ is in both $$G_{q}(\varepsilon)$$ and $$G_{r}(\varepsilon)$$, and such an $$x$$ exists since $$A \cap G_{\neg p}(\delta) \neq \emptyset$$ by assumption.

But this is impossible since $$G_{q}(\varepsilon) \cap G_{r}(\varepsilon)=\emptyset$$ $$(\text { a contradiction!). } \square$$

For intervals, see Chapter 3, §14, Example ( $$\mathrm{h} )$$.

corollary $$\PageIndex{2}$$

$$f$$ is continuous at $$p\left(p \in D_{f}\right)$$ iff $$f(x) \rightarrow f(p)$$ as $$x \rightarrow p$$.

Proof

The straightforward proof from definitions is left to the reader.

Note 3. In formula $$(2),$$ we excluded the case $$x=p$$ by assuming that $$x \in A \cap G_{\neg p}(\delta) .$$ This makes the behavior of $$f$$ at $$p$$ itself irrelevant. Thus for the existence of a limit $$q$$ at $$p,$$ it does not matter whether $$p \in D_{f}$$ or whether $$f(p)=q .$$ But both conditions are required for continuity at $$p$$ (see Corollary 2 and Definition 1$$)$$.

Note 4. Observe that if $$(1)$$ or $$(2)$$ holds for some $$\delta,$$ it certainly holds for any $$\delta^{\prime} \leq \delta .$$ Thus we may always choose $$\delta$$ as small as we like. Moreover, as $$x$$ is limited to $$G_{p}(\delta),$$ we may disregard, or change at will, the function values $$f(x)$$ for $$x \notin G_{p}(\delta)$$ ("local character of the limit notion").

II. Limits in E*. If $$S$$ or $$T$$ is $$E^{*}\left(\text { or } E^{1}\right),$$ we may let $$x \rightarrow \pm \infty$$ or $$f(x) \rightarrow \pm \infty .$$ For a precise definition, we rewrite $$(2)$$ in terms of $$globes$$ $$G_{p}$$ and $$G_{q} :$$

$\left(\forall G_{q}\right)\left(\exists G_{p}\right)\left(\forall x \in A \cap G_{\neg p}\right) \quad f(x) \in G_{q}.$

This makes sense also if $$p=\pm \infty$$ or $$q=\pm \infty .$$ We only have to use our conventions as to $$G_{ \pm \infty},$$ or the metric $$\rho^{\prime}$$ for $$E^{*},$$ as explained in Chapter 3, §11.

For example, consider

$^{\prime \prime}f(x) \rightarrow q \text{ as } x \rightarrow+\infty^{\prime \prime}\left(A \subseteq S=E^{*}, p=+\infty, q \in\left(T, \rho^{\prime}\right)\right).$

Here $$G_{p}$$ has the form $$(a,+\infty], a \in E^{1},$$ and $$G_{\neg p}=(a,+\infty),$$ while $$G_{q}=G_{q}(\varepsilon)$$, as usual. Noting that $$x \in G_{\neg p}$$ means $$x>a\left(x \in E^{1}\right),$$ we can rewrite $$\left(2^{\prime}\right)$$ as

$(\forall \varepsilon>0)\left(\exists a \in E^{1}\right)(\forall x \in A | x>a) \quad f(x) \in G_{q}(\varepsilon), \text{ or } \rho^{\prime}(f(x), q)<\varepsilon.$

This means that $$f(x)$$ becomes arbitrarily close to $$q$$ for large $$x(x>a)$$.

Next consider $$^{4} f(x) \rightarrow+\infty$$ as $$x \rightarrow-\infty$$ " Here $$G_{\neg p}=(-\infty, a)$$ and $$G_{q}=(b,+\infty] .$$ Thus formula $$\left(2^{\prime}\right)$$ yields (with $$S=T=E^{*},$$ and $$x$$ varying over $$E^{\mathrm{i}} )$$

$\left(\forall b \in E^{1}\right)\left(\exists a \in E^{1}\right)(\forall x \in A | x<a) \quad f(x)>b;$

similarly in other cases, which we leave to the reader.

Note 5. In $$(3),$$ we may take $$A=N$$ (the naturals). Then $$f : N \rightarrow\left(T, \rho^{\prime}\right)$$ is a sequence in $$T .$$ Writing $$m$$ for $$x,$$ set $$u_{m}=f(m)$$ and $$a=k \in N$$ to obtain

$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad u_{m} \in G_{q}(\varepsilon) ; \text{ i.e., } \rho^{\prime}\left(u_{m}, q\right)<\varepsilon.$

This coincides with our definition of the limit $$q$$ of a sequence $$\left\{u_{m}\right\}$$ (see Chapter 3, §14). Thus limits of sequences are a special case of function limits. Theorems on sequences can be obtained from those on functions $$f : A \rightarrow\left(T, \rho^{\prime}\right)$$ by simply taking $$A=N$$ and $$S=E^{*}$$ as above.

Note 6. Formulas $$(3)$$ and $$(4)$$ make sense sense also if $$S=E^{1}$$ (respectively, $$S=T=E^{1} )$$ since they do not involve any mention of $$\pm \infty .$$ We shall use such formulas also for functions $$f : A \rightarrow T,$$ with $$A \subseteq S \subseteq E^{1}$$ or $$T \subseteq E^{1},$$ as the case may be.

III. Relative Limits and Continuity. Sometimes the desired result $$(1)$$ or $$(2)$$ does not hold in full, but only with $$A$$ replaced by a smaller set $$B \subseteq A$$. Thus we may have

$(\forall \varepsilon>0)(\exists \delta>0)\left(\forall x \in B \cap G_{\neg p}(\delta)\right) \quad f(x) \in G_{q}(\varepsilon).$

In this case, we call $$q$$ a relative limit of $$f$$ at $$p$$ over $$B$$ and write

$"f(x) \rightarrow q \text{ as } x \rightarrow p \text{ over } B"$

or

$\lim _{x \rightarrow p, x \in B} f(x)=q \quad(\text { if } q \text { is unique });$

$$B$$ is called the path over which $$x$$ tends to $$p .$$ If, in addition, $$p \in D_{f}$$ and $$q=f(p),$$ we say that $$f$$ is relatively continuous at $$p$$ over $$B ;$$ then $$(1)$$ holds with $$A$$ replaced by $$B$$ . Again, if this holds for every $$p \in B,$$ we say that $$f$$ is relatively continuous on $$B .$$ Clearly, if $$B=A=D_{f},$$ this yields ordinary (nonrelative) limits and continuity. Thus relative limits and continuity are more general.

Note that for limits over a path $$B, x$$ is chosen from $$B$$ or $$B-\{p\}$$ only. Thus the behavior of $$f$$ outside $$B$$ becomes irrelevant, and so we may arbitrarily redefine $$f$$ on $$-B .$$ For example, if $$p \notin B$$ but $$\lim _{x \rightarrow p, x \in B} f(x)=q$$ exists, we may define $$f(p)=q,$$ thus making $$f$$ relatively continuous at $$p(\text { over } B) .$$ We also may replace $$(S, \rho)$$ by $$(B, \rho)(\text { if } p \in B),$$ or restrict $$f$$ to $$B,$$ i.e., replace $$f$$ by the function $$g : B \rightarrow\left(T, \rho^{\prime}\right)$$ defined by $$g(x)=f(x)$$ for $$x \in B$$ (briefly, $$g=f$$ on $$B )$$.

A particularly important case is

$A \subseteq S \subseteq E^{*}, \text{ e.g., } S=E^{1}.$

Then inequalities are defined in $$S,$$ so we may take

$B=\{x \in A | x<p\} \text{ (points in } A, \text{ preceding } p).$

Then, writing $$G_{q}$$ for $$G_{q}(\varepsilon)$$ and $$a=p-\delta,$$ we obtain from formula $$(2)$$

$\left(\forall G_{q}\right)(\exists a<p)(\forall x \in A | a<x<p) \quad f(x) \in G_{q}.$

If $$(5)$$ holds, we call $$q$$ a left limit of $$f$$ at $$p$$ and write

$"f(x) \rightarrow q \text{ as } x \rightarrow p^{-}" \quad\left(" x \text { tends to } p \text{ from the left}^{\prime}\right).$

If, in addition, $$q=f(p),$$ we say that $$f$$ is left continuous at $$p .$$ Similarly, taking

$B=\{x \in A | x>p\},$

we obtain right limits and continuity. We write

$f(x) \rightarrow q \text{ as } x \rightarrow p^{+}$

iff $$q$$ is a right limit of $$f$$ at $$p,$$ i.e., if $$(5)$$ holds with all inequalities reversed.

If the set $$B$$ in question clusters at $$p,$$ the relative limit (if any) is unique. We then denote the left and right limit, respectively, by $$f\left(p^{-}\right)$$ and $$f\left(p^{+}\right),$$ and we write

$\lim _{x \rightarrow p^{-}} f(x)=f\left(p^{-}\right) \text{ and } \lim _{x \rightarrow p^{+}} f(x)=f\left(p^{+}\right).$

corollary $$\PageIndex{3}$$

With the previous notation, if $$f(x) \rightarrow q$$ as $$x \rightarrow p$$ over a path $$B,$$ and also over $$D,$$ then $$f(x) \rightarrow q$$ as $$x \rightarrow p$$ over $$B \cup D$$.

Hence if $$D_{f} \subseteq E^{*}$$ and $$p \in E^{*},$$ we have

$q=\lim _{x \rightarrow p} f(x) \text{ iff } q=f\left(p^{-}\right)=f\left(p^{+}\right) . \quad(\text { Exercise! })$

We now illustrate our definitions by a diagram in $$E^{2}$$ representing a function $$f : E^{1} \rightarrow E^{1}$$ by its graph, i.e., points $$(x, y)$$ such that $$y=f(x)$$.

Here

$G_{q}(\varepsilon)=(q-\varepsilon, q+\varepsilon)$

is an interval on the $$y$$ -axis. The dotted lines show how to construct an interval

$(p-\delta, p+\delta)=G_{p}$

on the $$x$$ -axis, satisfying formula $$(1)$$ in Figure $$13,$$ formulas $$(5)$$ and $$(6)$$ in Figure $$14,$$ or formula $$(2)$$ in Figure $$15 .$$ The point $$Q$$ in each diagram belongs to the graph; i.e., $$Q=(p, f(p)) .$$ In Figure $$13, f$$ is continuous at $$p(\text { and also at }$$ $$p_{1}$$ ). However, it is only left-continuous at $$p$$ in Figure $$14,$$ and it is discontinuous at $$p$$ in Figure $$15,$$ though $$f\left(p^{-}\right)$$ and $$f\left(p^{+}\right)$$ exist. (Why?)

Example $$\PageIndex{1}$$

(a) Let $$f : A \rightarrow T$$ be constant on $$B \subseteq A ;$$ i.e.

$f(x)=q \text{ for a fixed } q \in T \text{ and all } x \in B.$

Then $$f$$ is relatively continuous on $$B,$$ and $$f(x) \rightarrow q$$ as $$x \rightarrow p$$ over $$B,$$ at each $$p .$$ (Given $$\varepsilon>0,$$ take an arbitrary $$\delta>0$$ . Then

$\left(\forall x \in B \cap G_{\neg p}(\delta)\right) \quad f(x)=q \in G_{q}(\varepsilon),$

as required; similarly for continuity.)

(b) Let $$f$$ be the $$i$$ dentity map on $$A \subset(S, \rho) ;$$ i.e.,

$(\forall x \in A) \quad f(x)=x.$

Then, given $$\varepsilon>0,$$ take $$\delta=\varepsilon$$ to obtain, for $$p \in A$$,

$\left(\forall x \in A \cap G_{p}(\delta)\right) \quad \rho(f(x), f(p))=\rho(x, p)<\delta=\varepsilon.$

Thus by $$(1), f$$ is continuous at any $$p \in A,$$ hence on $$A$$.

(c) Define $$f : E^{1} \rightarrow E^{1}$$ by

$f(x)=1 \text{ if } x \text{ is rational, and } f(x)=0 \text{ otherwise.}$

(This is the Dirichlet function, so named after Johann Peter Gustav Lejeune Dirichlet.)

No matter how small $$\delta$$ is, the globe

$G_{p}(\delta)=(p-\delta, p+\delta)$

(even the deleted globe) contains both rationals and irrationals. Thus as $$x$$ varies over $$G_{\neg p}(\delta), f(x)$$ takes on both values, 0 and $$1,$$ many times and so gets out of any $$G_{q}(\varepsilon),$$ with $$q \in E^{1}, \varepsilon<\frac{1}{2}$$.

Hence for any $$q, p \in E^{1},$$ formula $$(2)$$ fails if we take $$\varepsilon=\frac{1}{4},$$ say. Thus $$f$$ has no limit at any $$p \in E^{1}$$ and hence is discontinuous everywhere! However, $$f$$ is relatively continuous on the set $$R$$ of all rationals by Example $$(\mathrm{a})$$.

(d) Define $$f : E^{1} \rightarrow E^{1}$$ by

$f(x)=[x](=\text { the integral part of } x ; \text { see Chapter } 2, §10).$

Thus $$f(x)=0$$ for $$x \in[0,1), f(x)=1$$ for $$x \in[1,2),$$ etc. Then $$f$$ is discontinuous at $$p$$ if $$p$$ is an integer (why?) but continuous at any other $$p\left(\text { restrict } f \text { to a small } G_{p}(\delta) \text { so as to make it constant) }\right.$$

However, left and right limits exist at each $$p \in E^{1},$$ even if $$p=$$ $$n(\text { an integer }) .$$ In fact,

$f(x)=n, x \in(n, n+1)$

and

$f(x)=n-1, x \in(n-1, n),$

hence $$f\left(n^{+}\right)=n$$ and $$f\left(n^{-}\right)=$$ $$n-1 ; f$$ is right continuous on $$E^{1} .$$ See Figure $$16 .$$

(e) Define $$f : E^{1} \rightarrow E^{1}$$ by

$f(x)=\frac{x}{|x|} \text{ if } x \neq 0, \text{ and } f(0)=0.$

(This is the so-called signum function, often denoted by sgn.)

Then (Figure 17$$)$$

$f(x)=-1 \text{ if } x<0$

and

$f(x)=1 \text{ if } x>0.$

Thus, as in ( d ), we infer that $$f$$ is discontinuous at $$0,$$ but continuous at each $$p \neq 0 .$$ Also, $$f\left(0^{+}\right)=1$$ and $$f\left(0^{-}\right)=-1 .$$ Redefining $$f(0)=1$$ or $$f(0)=-1,$$ we can make $$f$$ right (respectively, left) continuous at $$0,$$ but not both.

(f) Define $$f : E^{1} \rightarrow E^{1}$$ by (see Figure 18$$)$$

$f(x)=\sin \frac{1}{x} \text{ if } x \neq 0, \text{ and } f(0)=0.$

Any globe $$G_{0}(\delta)$$ about 0 contains points at which $$f(x)=1,$$ as well as those at which $$f(x)=-1$$ or $$f(x)=0$$ (take $$x=2 /(n \pi)$$ for large integers $$n )$$; in fact, the graph "oscillates" infinitely many times between $$-1$$ and $$1 .$$ Thus by the same argument as in $$(\mathrm{c}), f$$ has no limit at 0 (not even a left or right limit) and hence is discontinuous at $$0 .$$ No attempt at redefining $$f$$ at 0 can restore even left or right continuity, let alone ordinary continuity, at $$0 .$$

(g) Define $$f : E^{2} \rightarrow E^{1} \mathrm{by}$$

$f(\overline{0})=0 \text{ and } f(\overline{x})=\frac{x_{1} x_{2}}{x_{1}^{2}+x_{2}^{2}} \text{ if } \overline{x}=\left(x_{1}, x_{2}\right) \neq \overline{0}.$

Let $$B$$ be any line in $$E^{2}$$ through $$\overline{0},$$ given parametrically by

$\overline{x}=t \vec{u}, \quad t \in E^{1}, \vec{u} \text{ fixed (see Chapter 3, §§4-6 ),}$

so $$x_{1}=t u_{1}$$ and $$x_{2}=t u_{2} .$$ As is easily seen, for $$\overline{x} \in B, f(\overline{x})=f(\overline{u})$$ (constant) if $$\overline{x} \neq \overline{0} .$$ Hence

$\left(\forall \overline{x} \in B \cap G_{\neg \overline{0}}(\delta)\right) \quad f(\overline{x})=f(\overline{u}),$

i.e., $$\rho(f(\overline{x}), f(\overline{u}))=0<\varepsilon,$$ for any $$\varepsilon>0$$ and any deleted globe about $$\overline{0}$$.

By $$\left(2^{\prime}\right),$$ then, $$f(\overline{x}) \rightarrow f(\overline{u})$$ as $$\overline{x} \rightarrow \overline{0}$$ over the path $$B .$$ Thus $$f$$ has a relative limit $$f(\overline{u})$$ at $$\overline{0},$$ over any line $$\overline{x}=t \overline{u},$$ but this limit is different for various choices of $$\overline{u},$$ i.e., for different lines through $$\overline{0} .$$ No ordinary limit at $$\overline{0}$$ exists (why?); $$f$$ is not even relatively continuous at $$\overline{0}$$ over the line $$\overline{x}=t \vec{u}$$ unless $$f(\overline{u})=0$$ (which is the case only if the line is one of the coordinate axes (why?)).