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Continuous functions, confusion with notation.

  1. Oct 21, 2005 #1
    My question reads:

    Let [itex]f[/itex] be defined and continuous on the interval [itex]D_1 = (0, 1)[/itex],

    and [itex]g[/itex] be defined and continuous on the interval [itex]D_2 = (1, 2)[/itex].

    Define [itex]F(x)[/itex] on the set [itex]D=D_1 \cup D_2 =(0, 2) \backslash \{1\}[/itex] by the formula:

    [tex]F(x)=f(x)[/tex], [tex]x\in (0, 1)[/tex]

    [tex]F(x)=g(x)[/tex], [tex]x\in (1, 2)[/tex]

    Is [itex]F[/itex] continuous or discontinuous?

    My confusion is with the notation, [itex]D=D_1 \cup D_2 =(0, 2) \backslash \{1\}[/itex].

    Does the [itex](0, 2) \backslash \{1\}[/itex] mean that set [itex]D[/itex] does not include the number 1?

    Also, if the above is true wouldn't that mean that [itex]F[/itex] is discontinuous?
  2. jcsd
  3. Oct 22, 2005 #2


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    Homework Helper

    (0,2)\{1} is indeed the set of all x such that
    0<x<2 and x!=1
    as is clear from its definition by union
    1 is in neither set so 1 is not in the union
    the function F is continuous
    This is a bit dependent on the exact definition used I presume

    A function f:R->R (or some subset there of) is continuous on a set S if
    for any x in S and h>0 there exist d(h,x)>0 such that
    |f(x)-f(a)|<h for all x in S that satisfy |x-a|<d(h,x)

    (using this definition a function is continuous at isolated points, ie the function f:{1,2}->{1,2} f(1)=2 f(2)=1 is continuous on {1,2} as |f(x)-1|<h for all points x such that x is in S and |x-1|<.999 as the only such value is x=1 likewise for x=2)

    Thus for F(x) consider a in D
    either f of g is continuous at the point, and F is defined using f and g
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