Is every well-defined map continuous?  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. 
 A: 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 $x = 1$.
A: Maybe I am misunderstanding something, but that function is clearly NOT continuous at $x=1$ just by definition of continuity: $F(1)=1$, but $\lim_{x\rightarrow1^-}F(x)=\frac{1}{2}$.
A: 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 $f(1)= \lim_{x\to1}f(x)$.
A: 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.   
