I've been given the following question, which I think I've completed, but I just wanted to check whether what I've said is valid.
Suppose that a PSTS(23) with a $K_5$ leave is constructed using the method outlined in the proof of Lemma 48, and then a maximum PSTS of order 23 is created by adding the triples $\{(0,0), (0,2), \infty_1\}$ and $\{(0,1),(0,2),\infty_2\}$. Determine the leave of the PSTS(22) obtained from this system by deleting the point $(0,2)$. Is this PSTS(22) a maximum PSTS(22)?
Without wanting to type out the entire lemma, we basically take a STS on 21 points, by taking the orbits of the following; $$B_1 = \{(0,0), (0,1), (0,2)\}$$ $$B_0 = \{\{(1,0), (6,0), (0,1) \}, \{(2,0), (5,0), (0,1)\}, \{(3,0), (4,0), (0,1)\}\}$$ Then, we basically delete the $B_1$ triple and all of its orbits, and set $\pi$ to be the permutation $(0)(1,2,...,6)$, and replace each triple of the form $\{(x,2), (y,2), (z,0)\}$ with $\{(x,2), (y,2), (\pi(z),0)\}$. Then, we have a PSTS of order 21, with a leave consisting of a 3-cycle and an 18-cycle. Letting the 18-cycle be $(x_1, x_2, ..., x_18)$, we add two points, $\infty_1$ and $\infty_2$, and add the triples $\{\infty_1, x_i, x_{i+1}\}$ for $i = 1,3,...,17$, $\{\infty_2, x_i, x_{i+1}\}$ for $i = 2,4,...,16$, and $\{\infty_1, x_1, x_18\}$. As a result of this, we basically get left with a PSTS on 23 points, with a $K_5$ leave on the points $\{(0,0), (0,1), (0,2), \infty_1, \infty_y\}$.
Now, adding in the triples $\{(0,0), (0,2), \infty_1\}$ and $\{(0,1),(0,2),\infty_2\}$, we reduce the $K_5$ leave to a $C_4$ leave, on the points $\{(0,0), (0,1), \infty_1, \infty_2\}$.
By one of the theorems given to us in class, this PSTS(23) will have a total of 83 triples.
Now, to find out what gets left if we delete the point $(0,2)$, we need to find out all the triples that contain $(0,2)$. By looking at a few of the remaining orbits and that, I found a total of 9 such triples, along with the 2 addition triples that we needed to add in. I'm not really keen on writing out all of these, but the removal of the point $(0,2)$ essentially removes all these triples, and we're left with a $C_4$, and 11 $K_2$ copies. So, that essentially answers the first question, yes??
Further, we have an additional theorem that tells us that a PSTS(22) will have a total of 73 triples, and have a leave of 9 $K_2$ graphs, and a $K_{1,3}$. The removal of $(0,2)$ gets rid of 11 triples, so the PSTS(22) will have 72 triples, and thus won't be maximum. We also know that the leave doesn't match. So, then, the PSTS(22) is not maximum.
I know this is a super long question, and I might have made a few mistakes, but if anyone could confirm what I've written here, I'd be very grateful. :)