Doubt about uniqueness of limit proof My lecturer gave us a proof of uniqueness of limit by contradiction, the context is: let $X \subset \mathbb{R}$ be an open set, let $x_0$ be a limit point for $X$ and let $f:X \to \mathbb{R}$ a function. If the limit of $f$ exists then it must be unique.
I have a doubt in the proof, which is the following: suppose by the sake of contradiction that $f$ has two limits $y_1 \ne y_2$ as $x \to x_0$. Let $V_1$ be a neighbourhood of $y_1$ and $V_2$ be a neighbourhood of $y_2$, with $V_1 \cap V_2 = \emptyset$. Since $x_0$ is a limit point, we can find two neighbourhoods $U_1$ and $U_2$ of the point $x_0$ such that for all $x \in U_1 \cap X \setminus \{x_0\}$ it is $f(x) \in V_1$ and for all $x \in U_2 \cap X \setminus \{x_0\}$ it is $f(x) \in V_2$. Since $U-1$ and $U_2$ are neighbourhoods of $x_0$ it is $U_1 \cap U_2 \ne \emptyset$ and since $x_0$ is a limit point there exists at least a point $\bar{x} \in X$ such that $\bar{x} \in U_1 \cap U_2$. So if $x \in U_1 \cap U_2$ it is $f(x) \in V_1$ and $f(x) \in V_2$, meaning $V_1 \cap V_2 \ne \emptyset$ and this contradicts the fact that $V_1 \cap V_2 = \emptyset$.
My doubt is: what if $V_1$ and $V_2$ aren't disjoint? I will try to explain my doubt: proofs must be general, and it seems to me that when he says &quot;let $V_1$ and $V_2$ be such that $V_1 \cap V_2 =\emptyset$&quot; we are restricting to a case that it isn't general because it imposes a condition on $V_1$ and $V_2$, that is that they are disjoint. Why this proof works then? I have a suspect that it is because this restriction is &quot;apparent&quot;, in the sense that we can always lead back to the case when $V_1$ and $V_2$ are disjoint (for example, if $V_1$ and $V_2$ aren't disjoint I can consider another neighbourhoods $V_1'$ and $V_2'$ with radius, for example, a fraction the radius of $V_1$ and $V_2$ big enough to assure that $V_1'$ and $V_2'$ are disjoint), and so obtain a general proof because the condition we impose can always be imposed and this makes $V_1 \cap V_2 =\emptyset$ an assumption without loss of generality. This has some sense or am I missing the point? Thanks.
I think that you are misunderstanding the meaning of the idea that proofs must be general. We are assuming that both $y_1$ and $y_2$ are limits of $f$ at $x_0$. Asserting that $y_1$ is limit of $f$ at $x_0$ means that for every neighborhood $V_1$ of $y_1$, there is a neighborhood $U_1$ of $x_0$ such that $x\in U_1\implies f(x)\in V_1$. And asserting that $y_2$ is limit of $f$ at $x_0$ means that for every neighborhood $V_2$ of $y_2$, there is a neighborhood $U_2$ of $x_0$ such that $x\in U_2\implies f(x)\in V_2$. Since this is true for every two neighborhoods $V_1$ and $V_2$ of $y_1$ and $y_2$ respectively, then, in particular, it must still be true when we choose those neighborhoods in such a way that they don't intersect. And doing that leads to a contradiction.
A proof being general means that it has to apply to every single case. But, in this context, this does not mean that it must work if we pick any two neighborhoods $V_1$ and $V_2$. If one pair of such neighborhoods leads to a contradiction, that's fine and we have then a proof.
I think that you are misunderstanding the meaning of the idea that proofs must be general. We are assuming that both $y_1$ and $y_2$ are limits of $f$ at $x_0$. Asserting that $y_1$ is limit of $f$ at $x_0$ means that for every neighborhood $V_1$ of $y_1$, there is a neighborhood $U_1$ of $x_0$ such that $x\in U_1\implies f(x)\in V_1$. And asserting that $y_2$ is limit of $f$ at $x_0$ means that for every neighborhood $V_2$ of $y_2$, there is a neighborhood $U_2$ of $x_0$ such that $x\in U_2\implies f(x)\in V_2$. Since this is true for every two neighborhoods $V_1$ and $V_2$ of $y_1$ and $y_2$ respectively, then, in particular, it must still be true when we choose those neighborhoods in such a way that they don't intersect. And doing that leads to a contradiction.
A proof being general means that it has to apply to every single case. But, in this context, this does not mean that it must work if we pick any two neighborhoods $V_1$ and $V_2$. If one pair of such neighborhoods leads to a contradiction, that's fine and we have then a proof.