Main derivation
S01 · implies
Before
\tau_X=\mathcal P(X),\quad\varnothing\subset\{u\}\subset YAfter
\tau_X,\tau_Y\text{ are topologies}Conditions: none
Rule: topology · Uses: G1
Leaf basis: topology
The source is the power set and the target a finite chain with endpoints; both satisfy every axiom.
S02 · implies
Before
f^{-1}(\varnothing),f^{-1}(\{u\}),f^{-1}(Y)After
\varnothing,X,X\in\tau_X
Conditions: none
Rule: definition_here · Uses: G1, G2, S01, definition_here
Leaf basis: definition_here
Check preimages of all target opens, proving continuity of f.
S03 · implies
Before
K=X=\{a,b\}After
K\text{ is compact}Conditions: none
Rule: definition_here · Uses: G1, definition_here
Leaf basis: definition_here
From any cover choose one original member for a and one for b; at most two cover K.
S04 · implies
Before
f(K)=\{u\}\subseteq\bigcup_{i\in I}W_i,\quad W_i\in\tau_YAfter
K\subseteq\bigcup_{i\in I}(K\cap f^{-1}(W_i))Conditions: none
Rule: sets · Uses: G2, S02
Leaf basis: sets
Take an arbitrary ambient-open cover of the image; each x in K has its image in some W_i, so the preimages cover K.
S05 · implies
Before
W_i\in\tau_Y
After
K\cap f^{-1}(W_i)\text{ is relatively open in }KConditions: none
Rule: definition_here · Uses: S02, S04, definition_here
Leaf basis: definition_here
Continuity gives an ambient-open preimage; intersection with K is relatively open.
S06 · theorem_application
Before
K\subseteq\bigcup_{i\in I}(K\cap f^{-1}(W_i))After
K\subseteq\bigcup_{i\in J}f^{-1}(W_i),\quad J\subseteq I\text{ finite}Conditions: none
Rule: definition_here · Uses: S03, S04, S05, definition_here
Leaf basis: definition_here
Compactness of K supplies a finite subset J of original indices.
S07 · implies
Before
z\in f(K),\ z=f(x),\ x\in K
After
z\in\bigcup_{i\in J}W_iConditions: none
Rule: sets · Uses: S06
Leaf basis: sets
Some selected preimage contains x, so the same W_i contains z; return exactly the same indices.
S08 · implies
Before
\forall(W_i)\text{ covering }f(K),\ \exists J\text{ finite covering }f(K)After
f(K)=\{u\}\text{ is compact but not closed in }YConditions: none
Rule: definition_here · Uses: S04, S07, definition_here
Leaf basis: definition_here
The first part proves compactness; {u} is not closed because {v} is not open, protecting the missing-Hausdorff boundary.