Are these alternative definitions of the "median" correct?

Question

(Textbook accepted) Definition: The median of a random variable X is any real value $m$ s.t. $P(X \geq m)\geq\frac{1}{2}$ and $P(X \leq m)\geq \frac{1}{2}$.

I've come up with 4 alternative definitions of the "median." I'm writing to ask whether those would also work against all counterexamples.

$(1)$

A "median" of a random variable X is any real value $m$ s.t. $P(X \geq m) < \frac{1}{2}$ and $P(X \leq m) < \frac{1}{2}$.

$(2)$

A "median" of a random variable X is any real value $m$ s.t. $P(X \geq m) < \frac{1}{2}$ and $P(X \leq m) \leq \frac{1}{2}$ .

$(3)$ A "median" of a random variable X is any real value $m$ s.t. $P(X \geq m) \leq \frac{1}{2}$ and $P(X \leq m) \leq \frac{1}{2}$

$(4)$

A "median" of a random variable X is any real value $m$ s.t. $P(X \geq m) \leq \frac{1}{2}$ and $P(X \leq m) < \frac{1}{2}$

Answer 1

For every $m\in\mathbb R$ we have:$$P(X\geq m)+P(X\leq m)=P(X=m)+1\geq1\tag1$$Now draw conclusions for (1),(2) and (4).

Concerning (3):

Note that $(1)$ implies that the "median" defined this way is forced to take a value $m$ such that $P(X=m)=0$, so it will not always coincide with what is normally called the median.

Answer 2

I don't see a particular reason to accept asymmetric definitions. Then number (1) must be rejected as it will result in "no median" if the $\text{cdf}$ takes the value $\frac12$.