Wouldn’t Every Envelope be the Greatest Element of a given Subset?

I'm having some problems understanding Badiou's definition of the envelope, and of the operation of union. Any help with this will be appreciated.

In the course of developing the transcendental algebra (LW, Book II, Section 3, Subsection 6) Badiou stipulates that the envelope of any given subset of T (defined by a certain property P) must obey the following two properties:

- It is greater than or equal than all the elements q of T that have property P.
- It is lesser than or equal to every element t which, like itself, is greater than or equal to all the elements q with property P (least upper bound)

Immediately afterwards Badiou stipulates that given a two element set B (p, q) then an upper bound would be greater than p or q. Then he stipulates that the envelope of this set would be the smallest of all elements greater than or equal to both p and q. I’m confused about this; how is it possible for the envelope of any non-empty set not to be the greatest element of that set? We have the following:

(1) B = {q ε T / P(q)} (The definition of the subset of T defined by property P)
(2) µ = ∑{q ε T / P(q)} (Definition of the envelope)

The envelope µ must be defined by the following properties:

(3) µ ≥ {p / P(p)} .∩. µ ≤ [t / t = ≥{p / P(p)}] (the envelope of B is equal to the smallest upper bound for B)

We stipulate by hypothesis that B is not empty, i.e. at least one element belongs to it.

(4) (B ≠ Ø) ↔ (Eq) (q ε B) ∩ (Vp) (p ε B) ∩ ( p ≤ q) (B is non-empty and is the largest element of B)
(5) P(q) [By definition of B, (1)]

Then we stipulate by hypothesis that the envelope µ is greater than all q that belong to B.

(6) µ > {q ε B .↔. P(q)}

But since we know that at least one q belongs to B, which is the greatest element in B, while also is smaller than µ, then being equal to itself, q satisfies (3), i.e. it obeys the two conditions required for the envelope.

(3) ∩ (1) ∩ (6) --> (7) (q < µ) .∩. q ↔ [q ≥ {p / P(p)} .∩. ≤ [t / t = ≥ {p / P(p)}]]

So it appears that q is both an upper bound (since it is equal to all the elements of B, being the largest element of that set)and the minimum upper bound, since it is equal to the maximum member in B. But by the definition of the envelope as the minimum upper bound for B it appears as if

(3) ∩ (7) ∩ (6) --> (q = µ) ∩ (µ > q)

So we arrive, it would appear, to a formal contradiction.

How could the union of p and q be thus a third element, greater than both? Since in every non empty set, there will be an element which is the greatest, it would follow it would always be the envelope, it seems. I think what may be missing would be the additional restrictive clause whereby if a set has a maximal value, then the envelope will be that value. Put it obversely, if a set there is no maximal value, then the envelope of that set will the least upper bound (for Badiou the distinction between 'lesser' and 'minimum' does not seem to be important, since what we are measuring is always degrees of intensity of appearance). If so then maybe we could rewrite (6) as µ ≥ {q ε B .↔. P(q)} and all would be fine, since we would end up identifying µ with q.