Let $X_i$ be a sequence of $n$ iid real valued random variables with distribution $F$ and density $f$ with respect to lebesgue measure. Compute the density function of the two smallest values in a sample of size $n$.
My attempt
Plan to compute the distribution function and then differentiate this to obtain the density function.
$$\mathbb P(X_{(1)} \leq t_1, X_{(2)} \leq t_2) = \mathbb P(X_{(1)} \leq \min\{t_1, t_2\})$$
since in the case that $t_1 \leq t_2$, the event $\{X_{(1)} \leq t_1, X_{(2)} \leq t_2\}$ is equivalent to the event $\{X_{(1)} \leq t_1\}$ and in the case that $t_1 > t_2$ the event is equivalent to $\{X_{(1)} \leq t_2\}$.
Now, differentiating $\mathbb P(X_{(1)} \leq \min\{t_1, t_2\})$ with respect to $t_1$ and $t_2$ is $0$ when $t_1 \neq t_2$, but this can't be correct since I expect the density to be $0$ only when $t_2 < t_1$.
Where did I make an incorrect assumption?
Your error is that you cannot say that the events
$E_1:=(X_{(1)} \leq t_1, X_{(2)} \leq t_2)$ and $E_2:=(X_{(1)} \leq \min\{t_1, t_2\})$ are the same.
You can just say that $E_1 \subset E_2$ (meaning $E_1 \implies E_2$), but $E_2$ cannot imply $E_1$ ; therefore you haven't the corresponding equality of probabilities.