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4.1: Basic Definitions

  • Page ID
    19042
  • ( \newcommand{\kernel}{\mathrm{null}\,}\)

    We shall now consider functions whose domains and ranges are sets in some fixed (but otherwise arbitrary) metric spaces (S,ρ) and (T,ρ), respectively. We write

    f:A(T,ρ)

    for a function f with Df=A(S,ρ) and Df(T,ρ).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 pS. In particular, if pA=Df( so that f(p) is defined) we  may ask: Is it possible to make the function values f(x) as near as we like ("ε near") to f(p) by keeping x sufficiently close ( "close ) to p, i.e., inside some sufficiently small globe Gp(δ)? 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(T,ρ), with A(S,ρ), is said to be continuous at p iff pA and, moreover, for each ε>0 (no matter how small) there is δ>0 such that ρ(f(x),f(p))<ε for all xAGp(δ). In symbols,

    (ε>0)(δ>0)(xAGp(δ)){ρ(f(x),f(p))<ε, or f(x)Gf(p)(ε)

    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 pA (since f(p) is not defined).

    If (1) holds for each p in a set BA, 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 Gp(δ) in (1) by the set Gp(δ){p}, i.e., the globe without its center, denoted G¬p(δ) and called the deleted δ -globe about p. This is even necessary if pDf. Replacing f(p) in (1) by some qT, we then are led to the following definition.

    Definition

    Given f:A(T,ρ),A(S,ρ),pS, and qT, we say that f(x) tends to q as x tends to p(f(x)q as xp) iff for each ε>0 there is δ>0 such that ρ(f(x),q)<ε for all xAG¬p(δ). In symbols,

    (ε>0)(δ>0)(xAG¬p(δ)){ρ(f(x),q)<ε, i.e. f(x)Gq(ε)

    This means that f(x) is ε -close to q when x is δ -close to p and xp.

    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( for a given p), we write q=limxpf(x).

    Note 1. Formula (2) holds "vacuously" (see Chapter 1,8 §§1-3, end remark) if AG¬p(δ)= for some δ>0. Then any qT 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 AG¬p(δ) for all δ>0, as we prove below.

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

    (δ>0)AG¬p(δ).( Explain! )

    Thus we have the following corollary.

    corollary 4.1.1

    If A clusters at p in (S,ρ), then a function f:A(T,p) can have at most one limit at p; i.e.

    limxpf(x) is unique (if it exists).

    In particular, this holds if A(a,b)E1(a<b) and p[a,b].

    Proof

    Suppose f has two limits, q and r, at p. By the Hausdorff property,

    Gq(ε)Gr(ε)= for some ε>0.

    Also, by (2), there are δ,δ>0 such that

    (xAG¬p(δ))f(x)Gq(ε) and (xAG¬p(δ))f(x)Gr(ε)

    Let δ=min(δ,δ). Then for xAG¬p(δ),f(x) is in both Gq(ε) and Gr(ε), and such an x exists since AG¬p(δ) by assumption.

    But this is impossible since Gq(ε)Gr(ε)= ( a contradiction!). 

    For intervals, see Chapter 3, §14, Example ( h).

    corollary 4.1.2

    f is continuous at p(pDf) iff f(x)f(p) as xp.

    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 xAG¬p(δ). 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 pDf 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 δ, it certainly holds for any δδ. Thus we may always choose δ as small as we like. Moreover, as x is limited to Gp(δ), we may disregard, or change at will, the function values f(x) for xGp(δ) ("local character of the limit notion").

    II. Limits in E*. If S or T is E( or E1), we may let x± or f(x)±. For a precise definition, we rewrite (2) in terms of globes Gp and Gq:

    (Gq)(Gp)(xAG¬p)f(x)Gq.

    This makes sense also if p=± or q=±. We only have to use our conventions as to G±, or the metric ρ for E, as explained in Chapter 3, §11.

    For example, consider

    f(x)q as x+(AS=E,p=+,q(T,ρ)).

    Here Gp has the form (a,+],aE1, and G¬p=(a,+), while Gq=Gq(ε), as usual. Noting that xG¬p means x>a(xE1), we can rewrite (2) as

    (ε>0)(aE1)(xA|x>a)f(x)Gq(ε), or ρ(f(x),q)<ε.

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

    Next consider 4f(x)+ as x " Here G¬p=(,a) and Gq=(b,+]. Thus formula (2) yields (with S=T=E, and x varying over Ei)

    (bE1)(aE1)(xA|x<a)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(T,ρ) is a sequence in T. Writing m for x, set um=f(m) and a=kN to obtain

    (ε>0)(k)(m>k)umGq(ε); i.e., ρ(um,q)<ε.

    This coincides with our definition of the limit q of a sequence {um} (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(T,ρ) by simply taking A=N and S=E as above.

    Note 6. Formulas (3) and (4) make sense sense also if S=E1 (respectively, S=T=E1) since they do not involve any mention of ±. We shall use such formulas also for functions f:AT, with ASE1 or TE1, 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 BA. Thus we may have

    (ε>0)(δ>0)(xBG¬p(δ))f(x)Gq(ε).

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

    "f(x)q as xp over B"

    or

    limxp,xBf(x)=q( if q is unique );

    B is called the path over which x tends to p. If, in addition, pDf 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 pB, we say that f is relatively continuous on B. Clearly, if B=A=Df, 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 pB but limxp,xBf(x)=q exists, we may define f(p)=q, thus making f relatively continuous at p( over B). We also may replace (S,ρ) by (B,ρ)( if pB), or restrict f to B, i.e., replace f by the function g:B(T,ρ) defined by g(x)=f(x) for xB (briefly, g=f on B).

    A particularly important case is

    ASE, e.g., S=E1.

    Then inequalities are defined in S, so we may take

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

    Then, writing Gq for Gq(ε) and a=pδ, we obtain from formula (2)

    (Gq)(a<p)(xA|a<x<p)f(x)Gq.

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

    "f(x)q as xp"("x tends to p from the left).

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

    B={xA|x>p},

    we obtain right limits and continuity. We write

    f(x)q as xp+

    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(p) and f(p+), and we write

    limxpf(x)=f(p) and limxp+f(x)=f(p+).

    corollary 4.1.3

    With the previous notation, if f(x)q as xp over a path B, and also over D, then f(x)q as xp over BD.

    Hence if DfE and pE, we have

    q=limxpf(x) iff q=f(p)=f(p+).( Exercise! )

    We now illustrate our definitions by a diagram in E2 representing a function f:E1E1 by its graph, i.e., points (x,y) such that y=f(x).

    Here

    Gq(ε)=(qε,q+ε)

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

    (pδ,p+δ)=Gp

    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( and also at  p1 ). However, it is only left-continuous at p in Figure 14, and it is discontinuous at p in Figure 15, though f(p) and f(p+) exist. (Why?)

    Example 4.1.1

    (a) Let f:AT be constant on BA; i.e.

    f(x)=q for a fixed qT and all xB.

    Screen Shot 2019-06-03 at 11.55.40 PM.png

    Screen Shot 2019-06-03 at 11.56.01 PM.png

    Then f is relatively continuous on B, and f(x)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.

    Screen Shot 2019-06-03 at 11.58.51 PM.png

    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 .

    Screen Shot 2019-06-04 at 12.07.32 AM.png

    (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.

    Screen Shot 2019-06-04 at 12.13.52 AM.png

    (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 .

    Screen Shot 2019-06-04 at 12.18.29 AM.png

    (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?)).


    This page titled 4.1: Basic Definitions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.