What does it mean for a function to be defined on an interval?

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A function defined on an interval means it has a specific range of input values where it is applicable, without gaps in that range. An interval is a convex subset of real numbers, ensuring that any linear combination of points within it also lies within the interval. Discontinuities are allowed as long as there are no gaps in the defined set of values. For example, a piecewise function can be defined on an interval but may not be continuous. The essential aspect is that each input in the interval corresponds to a single output, adhering to the definition of a function.
AlfredPyo
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So I was looking at one of the definitions of first order DE's.
But I don't get what this statement means:

let a function f(x) be defined on an interval I.
 
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I assume you are working with real numbers.
A (real) interval is a convex subset of the real numbers.
So I is some subset of R such that
$$\text{if }x_k,a_k\in \mathbb{R} \text{ and }x_k\in I \text{ with}\sum_{k=1}^n a_k= 1 \text{ then }\sum_{k=1}^n x_k a_k \in I$$
In other words we have a block of values for which the function is defined for all valuesIn differential equations we can have problems if the function is defined on a set with gaps such as in the set of integers or rationals.

Things like
$$x\le 7\\
x>57\\
3<x\le 7 \\
x\in \mathbb{R}\\
x\in \emptyset\\
x=1$$
are what you should have in mind
 
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Ok, so basically to be defined means to not have any discontinuities, right?
 
^There can be discontinuities, there cannot be gaps.
$$\mathrm{f}(x)=\begin{cases}\phantom{\frac{0}{0}}0 & x<0 \\
\,\,\,\, \frac{1}{2} & x=0\\
\phantom{\frac{0}{0}}1&x>0\end{cases}$$
is defined on an interval, but not continuous
 
Ok, so a piecewise function. As long as in the interval, all x values in the interval has one y value or vice versa?
 
That's pretty much the definition of "function", isn't it?
 

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