I recently came accross this property about alpha (or lambda) cuts
$$ (A')_\alpha \neq (A_\alpha)', unless \\ \alpha = 0.5 $$ where A is a fuzzy set with membership function $\mu_A(x)$
I am curious about the equality condition which I tried proving it as follows:-
We know that
$$ \mu^{'}_{A}(x) = 1 - \mu_A(x) $$
Now
$$ A_\alpha = \{x | \mu_A(x)\geq \alpha \} $$
$$ \implies (A_\alpha)' = \{ x | \mu_A(x) < \alpha \} $$
Similarly
$$ (A')_\alpha = \{ x | 1 - \mu_A(x) \geq \alpha \} $$
Given that
$$ (A')_\alpha = (A_\alpha)' $$
We thus have
$$ \{ x | 1 - \mu_A(x) \geq \alpha \} = \{ x | \mu_A(x) < \alpha \} $$
$$ \implies 1 - \mu_A(x) \geq \alpha $$
And
$$ \mu_A(x) < \alpha $$
$$ \implies 1 - 2\alpha \geq 0 $$
$$ \implies \alpha \leq 0.5 $$
This is where I am stuck, how do I go about eliminating the inequality and obtain just $\alpha = 0.5$ ?
Thanks!