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Two-toss sample space 2026-10-14

Model two fair independent coin tosses, recording their order. Here

H

denotes heads and

T

tails, and

Ω = {HH, H T, T H, TT}, F = P (Ω), P ({ω}) = \frac{1}{4} (ω \in Ω) .

P (Ω)

includes every question expressible as a subset of these four outcomes. The mass

1/4 = (1/2) (1/2)

comes from the fairness and independence assumptions. These are exact model probabilities, rather than measured toss frequencies.

Let

A

be the event that the first toss is heads and

B

the event that the second toss is heads. Then

A = {HH, H T}, B = {HH, T H}, A \cup B = {HH, H T, T H}, A \cap B = {HH} .

The union means “at least one head,” while the intersection means “both heads.” Their probabilities are

3/4

and

1/4

. Counting the union through the inclusion-exclusion principle gives

1/2 + 1/2 - 1/4 = 3/4

; alternatively its complement of an event is

{TT}

, of probability

1/4

.

Figure 1.
Four ordered outcomes and the event of at least one head
. Rows record the first toss and columns the second. Each cell has probability $1/4$ under fair independent tosses. The highlighted union $A \cup B$ includes $HH$ , $H T$ , and $T H$ ; the overlap $A \cap B$ contains only $HH$ . Details.

On the same sample space, assigning masses

3/8, 1/8, 1/8, 3/8

to

HH, H T, T H, TT

would still make each toss fair:

P (A) = P (B) = 1/2

. But

P (A \cap B) = 3/8 \neq = 1/4

, so the tosses would be dependent. The event of at least one head would have probability

5/8

. Changing the probability measure changes the answer even when the sample space and the single-toss probabilities stay the same.

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Vector space bases from Zorn's lemma 2026-10-14

Every vector space

V

over a field

F

has a basis, assuming the axiom of choice. More generally, any linearly independent subset

A \subseteq V

extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing

A

:

P_{A} = {S \subseteq V : A \subseteq S, S is linearly independent over F}, S \leq T ⟺ S \subseteq T .

It is a set, being a subset of the power set of

V

, and is nonempty because

A \in P_{A}

.

A short proof outline supplies both essential steps:

For a nonempty chain in a partial order $C \subseteq P_{A}$ , put $U = ⋃_{S \in C} S$ . Any finite linear relation in $U$ lies in one member of $C$ : among finitely many containing members, one contains the rest. Its linear independence makes every coefficient zero. Thus $U \in P_{A}$ is an upper bound. For the empty chain in a partial order, use $A$ ; its empty union need not contain $A$ .
Zorn's lemma gives a maximal element $B$ . If a vector $v$ lies outside its linear span, then $B \cup {v}$ is linearly independent: a linear relation with nonzero coefficient on $v$ would express $v$ in that linear span. This contradicts maximality, so $B$ is a spanning set and hence a basis.^[ref]

Taking

A = \emptyset

proves existence, including

V = {0}

, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail:

{v}

and

{2 v}

for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

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Zermelo set theory 2026-10-14

The usual set theory with extensionality, empty set, pairing, union, power set, infinity and full separation, before adjoining replacement. Modern presentations often include foundation; the convention should be specified when comparing theories. Class-theoretic variants add classes and class comprehension, rather than treating proper classes as sets.

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