Continuous functions, confusion with notation.

[*melinda*]

hi,
My question reads:

Let $f$ be defined and continuous on the interval $D_1 = (0, 1)$,

and $g$ be defined and continuous on the interval $D_2 = (1, 2)$.

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

$$F(x)=f(x)$$, $$x\in (0, 1)$$

$$F(x)=g(x)$$, $$x\in (1, 2)$$

Is $F$ continuous or discontinuous?

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

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

Also, if the above is true wouldn't that mean that $F$ is discontinuous?

 

[lurflurf]

hi,
My question reads:
Let $f$ be defined and continuous on the interval $D_1 = (0, 1)$,
and $g$ be defined and continuous on the interval $D_2 = (1, 2)$.
Define $F(x)$ on the set $D=D_1 \cup D_2 =(0, 2) \backslash \{1\}$ by the formula:
$$F(x)=f(x)$$, $$x\in (0, 1)$$
$$F(x)=g(x)$$, $$x\in (1, 2)$$
Is $F$ continuous or discontinuous?
My confusion is with the notation, $D=D_1 \cup D_2 =(0, 2) \backslash \{1\}$.
Does the $(0, 2) \backslash \{1\}$ mean that set $D$ does not include the number 1?
Also, if the above is true wouldn't that mean that $F$ is discontinuous?

(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