How to conclude that for all $f,g \in S_I$, we have that if $f(x)=g(x)$ for some $x \in I,$ then $f=g$? Motivation. If we're trying to solve the differential equation $y'=y$ on the interval $I$, we would probably like to assume that for all $x \in I$ we have that $y(x) \neq 0.$ But, we need to justify this assumption by showing that, if we assume to the contrary that there exists $x \in I$ such that $y(x) = 0,$ then the only possible solution is $y=0_I$, where $0_I$ is the function with domain $I$ that is everywhere $0$. This motivates the following question.
Question. Suppose that for all sets $X \subseteq \mathbb{R},$ we define that $$S_X = \{y : X \rightarrow \mathbb{R} \mid y'=y;\; y \mbox{ diff}\}.$$
Under this definition, is there some theorem that allows us to conclude that for all non-trivial* intervals $I$ and for any two $f,g \in S_I$, we have that if $f(x)=g(x)$ for some $x \in I,$ then $f=g$?
*By a non-trivial interval, I mean an interval incorporating two or more elements (which implies that the interval has infinitely many elements). Actually, I'm not certain whether this assumption is strictly necessary, but it seems harmless.
 A: Let's take your interval $I$ and a point $x_0\in I$. Examine the Cauchy problem
$$\begin{cases}y'(x)=y(x),&x\in I\\y(x_0)=y_0.&{}\end{cases}$$
Clearly, it a has a unique solution in a neighbourhood of $(x_0,y_0)$. If you take a look at the proof of the Picard theorem, you notice that in our case the interval of existence and unicity doesn't depend on the initial data, hence we can prolongate our solution to whe whole interval $I$ without losing unicity.
Another approach would be to take two different solutions of that Cauchy problem, take their difference; it satisfies the system
$$\begin{cases}y'(x)=y(x),&x\in I\\y(x_0)=0.&{}\end{cases}$$
We multiply both parts by $e^{-x}$ to obtain $(y(x)e^{-x})'=0,\, e^{-x_0}y(x_0)=0$ which apparently has a unique solution which is identically zero.
Edit
Look at the detailed proof on wiki. The interval of existence $(x_0-a,x_0+a)$ is defined by constants $M$, $L$, $b$: $a<\min(b/M, 1/L)$. $M$ represents the sup-norm of right-hand side of ODE, $L$ - the Lipschitz constant for the right-hand side and $b$ is a certain constant that you fix before (the neighborhood of $y_0$).
Without losing generality, $I=[0,1]$ and $x_0=0$. We have $L=1$, let's take $b=e|y_0|$, hence $M=(e+1)|y_0|$. We get $$a<\min(b/M, 1/L)=\min \left(\frac{e}{e+1},1\right).$$ Note that the obtained interval doesn't depend on initial data. For example, let's take $a = 1/2$. The solution is unique on $[0,0.5]$ and therefore is equal to $y_0e^x$.
Now we study the Cauchy problem to continue the solution:
$$\begin{cases}y'(x)=y(x),&x\in [1/2,1]\\y(1/2)=y_0e^{1/2}.&{}\end{cases}$$
By applying the above reasoning we once again obtain that the interval of existence is bounded by the same $a< \min \left(\frac{e}{e+1},1\right)$. We can once again take $a=1/2$, so the the solution is unique on $[0,1]$ and thus is "glued" to the previous one.
We conclude that the solution is inuque on the whole $I$ and is equal to $y_0e^x$.
