I have 3 independent random variables, $X_i$, distributed on a continuous uniform distribution between 0 and 1.
Does the following hold given the assumptions above?
$$ \tag{1} P\left(X_1 < X_2 < X_3\right) = P\left(X_1 \le X_2 \le X_3\left|X_1\ne X_2\ne X_3\right.\right) $$
I call $A \equiv (X_1 < X_2 < X_3)$ and $B \equiv (X_1 \le X_2 \le X_3)$ and $P(N) \equiv P(X_1\ne X_2 \ne X_3)$ to simplify as, $$ \tag{2} P(A) = P(B|N) $$
The probability, of events not equalling each other, $P(N)$ is defined by,
$$ \tag{3} P(N) = 1 - P(X_1 = X_2 \cap X_2 < X_3) \cup P(X_1 < X_2 \cap X_2 = X_3) \cup P(X_1 =X_2=X_3) $$
I investigate the probability of $P(X = a)$ for the continuous uniform distribution, by taking the following limit $$ P\left(X=a\right) = \lim_{\epsilon\to0} P\left(a-\epsilon < X \le a\right) = \lim_{\epsilon\to0}\int^{a}_{a-\epsilon} dx = \lim_{\epsilon\to0} \epsilon = 0 $$ Thus, for a continuous uniform distribution between 0 and 1, $P(X_i=X_j)=\delta_{ij}$, so we get, $P(N) = 1$ for (3). We can transform (1) as follows,
$$ \tag{4} P(A) = \frac{P(B \cap N)}{P(N)} $$
because events $X_i$ are independent,
$$ \tag{5} P(A) = \frac{P(B)P(N)}{P(N)} = P(A) $$
Thus, $P(A) = P(B)$
Three independent and uniformly distributed continuous random variables will almost certainly not be coequal. That is, $\mathsf P(X_1\neq X_2, X_2\neq X_3)=1$
$$\therefore\mathsf P(X_1<X_2<X_3)=\mathsf P(X_1\leq X_2\leq X_3)$$