First decomposition theorem of fuzzy set

Question

I read if standard union in fuzzy set have definition:

Union of two fuzzy sets $\tilde{A}$ and $\tilde{B}$ in universe $X$ denoted $\tilde{A}\cup\tilde{B}$ is fuzzy set in universe $X$ with membership function $$ \mu_{\tilde{A}\cup\tilde{B}}(x)=max\{\mu_\tilde{A}(x),\mu_\tilde{B}(x)\} $$ But in prove of first decomposition theorem, I found this

$$ \mu_{\bigcup_{\alpha\in[0,1]}\tilde{A}_\alpha}(x)=\sup_{\alpha\in[0,1]}\mu_{\tilde{A}_\alpha}(x)\\ \mu_{\bigcup_{\alpha\in[0,1]}\tilde{A}_\alpha}(x)=\max\{\sup_{\alpha\in[0,a]}\mu_{\tilde{A}_\alpha}(x),\sup_{\alpha\in(a,1]}\mu_{\tilde{A}_\alpha}(x)\} $$

So why first using supremum to union of fuzzy set and then using maximum again?

If X is set with partial ordering relations, I know if $S\subseteq X$ $$ \sup S = \max S $$ But in the theorem just say $X$ is universe.

Note:

1. $\tilde{A}$ is fuzzy set in universe $X$.
2. $A_\alpha$ is strong $\alpha$-cut from $\tilde{A}$.
3. $\tilde{A}_\alpha$ is fuzzy set in universe $X$ with membership function. $$ \mu_{\tilde{A}_\alpha}(x)= \begin{cases} \alpha & \mbox{if $x\in A_\alpha$}\\ 0 & \mbox{if $x \notin A_\alpha$} \end{cases} $$