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I am working on the function $f:[0,1) \to [0,1]$ defined by $f(x)=\dfrac{x}{x+1}$ for each $x\in [0,1)$.

It is clear that $f$ is well defined and continuous.

Now consider another function given by $$ F(x)= \begin{cases} f(x), &x\in[0,1), \\ 1, & x = 1. \end{cases} $$ It is clear that $F$ is an extension of $f$ into $[0,1]$. But it is not clear to me that $F$ is not continuous at $x=1$, even though it is defined at $x=1$. Please I need more clarification about this.

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Do you know any definition of continuity adequate for metric spaces? How is continuity at 1 related to $lim_{x \to 1}f(x)$? –  savick01 Oct 30 '11 at 10:21
    
What is $\lim_{x\to1}f(x)$? –  AD. Oct 30 '11 at 10:21
    
@AD $\lim_{x\to 1}f(x)=\frac{1}{2}$ –  Hassan Muhammad Oct 30 '11 at 10:27
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I don't understand the question Does every well define map continuous? –  wildildildlife Oct 30 '11 at 12:38

4 Answers 4

up vote 4 down vote accepted

In general, a function $g$ is well-defined at $a$ if $g(a)$ exists.

In general, a function $g$ is continuous at $a$ if:

  • $g(a)$ exists, and
  • $\lim\limits_{x\to a} g(x)$ exists, and
  • $g(a) = \lim\limits_{x\to a}g(x)$.

Therefore, if a function $g$ is continuous at $a$, then it must be well-defined there. However, the converse may not be true: a function may be well-defined at a point, but not continuous there.

For example, your function $F$ is well-defined at $x = 1$ (since $F(1)$ exists), but is not continuous at $x = 1$ (since $\lim\limits_{x \to 1}F(x) \neq F(1)$.)

On the other hand, note that while your function $f$ is well-defined and continuous on $[0,1)$, it is not well-defined at $x = 1$, and so it is not continuous at $x = 1$.

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If I get you, every continuous function at a point $x_{0}$ is well define at that point, if the function is not well define at $x_{0}$ then it is not continuous there. –  Hassan Muhammad Oct 30 '11 at 16:28
    
@HassanMuhammad: Yes, that's right. –  Jesse Madnick Oct 30 '11 at 18:07

Maybe I am misunderstanding something, but that function is clearly NOT continuous at $x=1$ just by definition of continuity: $F(1)=1$, but $\lim_{x\rightarrow1^-}F(x)=\frac{1}{2}$.

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Which implies which? Well define or continuity? –  Hassan Muhammad Oct 30 '11 at 10:30
    
@Hassan The function $f$ is well-defined, but not continuous. If the limit of f at 1 is well-defined depends on your definition of limit. So, what is the object whose well-definedness concerns you? –  Phira Oct 30 '11 at 10:37
    
@Hassan: you should clarify yourself the meaning of well-definedness and continuity. Well-defined means that the function attains a particular value, without ambiguity - and this is true in your case. Continuity, assuming the Euclidean topology, means that the value $F(x_0)$ equals the limit $\lim_{x\rightarrow x_0}F(x)$. So, your function is well-defined but not continuous at $x=1$. –  Valerio Capraro Oct 30 '11 at 10:41
    
@ValerioCapraro: I agree with you. Jesse solution make sense, and I think you are saying the same thing. –  Hassan Muhammad Oct 30 '11 at 16:43
    
yes, we are saying the same thing. –  Valerio Capraro Oct 30 '11 at 17:15

As the variable $x$ tends to $1$, $F(x)$ tends to $1/2$, which is different from the value of $F$ at the point $1$. Intuitively, the graph of $F$ has more than one piece near the point 1, which is said to be discontinuous at the point $1$.

What should be noticed is that sometimes, even the graph has more than one piece near a point $x_{0}$, we still regard it a continuous function at $x_{0}$ as long as $\lim_{x\rightarrow x_0}f(x)=f(x_0)$. For example, the function

$G(x)=x \sin\frac{1}{x}$ for $x\neq 0, G(0)=0$

has three pieces near 0, but it's continuous at 0.

So the continuousness is just a description of the function having one piece near a point in the "limit" sense.

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We may always extend $f$ to a well-defined function on $[0,1]$ - to do this we just have to say what $f(1)$ is.

Since $\lim_{x\to1}f(x)$ exists we may extend $f$ to a continuous function on $[0,1]$ - to do this we must set $f(1)= \lim_{x\to1}f(x)$.

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This is a good exercise. Do you try proving? –  Hassan Muhammad Oct 30 '11 at 16:31
    
@DanPetersen You are totally right! I removed that. :) –  AD. Oct 30 '11 at 19:30

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