Download Proof Step Lists. Updated 3-3

PROOF-STEP-LIST FOR MATH 550
Introduction. The latest homework, and many of the extra credit
homework problems involve writing down correct proofs of mathematical assertions. In what follows we will give a list of steps in the proofs
that are required to prove different types of assertions. Carefully stated
definitions are a key in knowing what the steps are; another key is familiarity with basic logic.
Show that a specific set U is open. Recall the definition of an open
set: U ⊂ Rn is open if
(1) For every x ∈ U and
(2) there exists a number δ > 0 such that
(3) for every y ∈ Rn
(4) if ky − xk < δ
(5) then y ∈ U .
The steps of the proof are exactly in parallel to the above. Phrases
enclosed in quotes ‘. . . ’ should be written down verbatim in the proof,
changing only the names of the variables or unspecified constants to
conform to the notation in the assigned problem.
• ‘Let x ∈ U be given.’
• ‘Define δ’ in terms of x and U , and verify that δ > 0.
• ‘Let y ∈ Rn be given’, and
• ‘assume that ky − xk < δ.’
• Show that y ∈ U . Generally this argument involves both the
membership conditions of the set U as well as the specific value
of δ specified in the second step of the proof.
Show that U ⊂ V . Recall the definition: U ⊂ V means that
(1) For every x ∈ U
(2) it follows that x ∈ V .
Generally U = {x | P (x)}, where the membership condition P (x) is
true exactly when x ∈ U . Similarly V = {x | Q(x)}, where Q(x) is
the membership condition for the set V . The proof that U ⊂ V then
proceeds in the following steps.
• ‘Let x be given and assume P (x) is true’ (so that x ∈ U ).
• Show that Q(x) is true (i.e. x ∈ V ).
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Show that x is a boundary point of U . Recall the definition: x is
a boundary point of U ⊂ Rn if
(1)
(2)
(3)
(4)
(5)
(6)
For every δ > 0
there exists y, z ∈ Rn such that
ky − xk < δ and
kz − xk < δ and
y ∈ U and
z∈
/ U.
Consequently the proof of this proceeds in the following steps:
• ‘Let δ > 0 be given.’
• ‘Define y’ and ‘define z’ in terms of x and δ and U , and verify
that y, z ∈ Rn .
• Show that ky − xk < δ.
• Show that kz − xk < δ.
• Show that y ∈ U .
• Show that z ∈
/ U.
Show that “P if and only if Q”. Here P and Q are propositions
(mathematical assertions) and “P if and only if Q” is a compound
assertion formed from P and Q. The proof proceeds in the following
steps.
• ‘Assume P is true.’
• Show that Q is true. This proves “if P then Q”.
• ‘Assume Q is true.’ (No longer assume a priori that P is true
as in the first two steps.)
• Show that P is true. This proves “if Q then P ”.
An assertion of the form “if P then Q” is called a logical implication. We
have in the above advocated a direct proof of each of the implications “if
P then Q” and “if Q then P ”. An alternate proof (to the direct proof)
for any logical implication is the proof of the contrapositive. Such a
proof of “if P then Q” proceeds in the following steps.
• ‘Assume Q is not true.’
• Show that P is not true.
Proof of the contrapositive is sometimes much easier than a direct
proof. Any of the ‘show that’ steps in the above proof-step-lists might
be significantly expanded into further steps using the definitions involved in (and the logical structure of) the assertion whose truth is to
be shown. Similarly, the step ‘Assume the proposition P is true’ can be
expanded when the definition of each term involved in the assumption
P is applied.
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Applying Definitions. So much of the work in proofs is correct application of definitions. First of all this requires that one have a wellcrafted statement of the definition to be applied. A typical definition
involves a number of symbols, some which will have the same meaning in any application, and some others which might be interpreted
(or made specific) in a particular application. In some cases certain
symbols are overloaded, i.e. they can have different meanings in different occurences within the definition depending on the context. The
definition will usually involve various technical terms and/or notations
which have been previously defined. The specific technical term and/or
notation being defined usually applies only in a specific context, so part
of the definition is to spell out this context. To illustrate these aspects
we will give a careful statement of the definition of a linear mapping,
and follow it with a parsing of it into the elements we have mentioned
above.
Definition. Suppose V, + , ·, 0 and W, + ′ , ·′ , 0′ are vector spaces and
F : V → W is a mapping. F is linear if both of the following conditions
are true.
(1) For all v1 , v2 ∈ V we have that F (v1 + v2 ) = F (v1 ) + ′ F (v2 ).
(2) For all v ∈ V and all a ∈ R we have that F (v · a) = F (v) ·′ a.
Now here are the elements.
• Symbols which will have the same meaning in all applications:
: , →, ∈, =, R.
• Symbols which will need to be interpreted or made specific in
a given application: V, W, + , + ′ , ·, ·′ , 0, 0′ , F
• Overloaded symbols: none, because we used primes to distinguish + in V and + ′ in W, etc. Often definitions are not as
pedantic as this one is.
• Technical terms and/or notations previously defined: vector
space, mapping, F : V → W, F (v).
• Technical term to be defined: linear. Often the technical term
being defined is italicized.
• Context: the suppositions stated in the first sentence. It is
assumed that the reader knows that if V, +, ·, 0 and W, +′ , ·′ , 0′
are vector spaces then V and W are sets, so it is sensible to also
suppose that F : V → W is a mapping.
• The symbols v1 , v2 , v, a are dummy variables because they are
under universal quantifiers ‘for all’. Each of them could be
replaced by any other symbol (not already used) as long as all
occurences are replaced by the same symbol, and the meaning
of the condition would be unchanged.
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To illustrate how to apply this definition, and hopefully general principles concerning how most any definition would be applied, we will
perform the following exercise.
Problem. Suppose V, + , ·, 0 is the vector space discussed in the addendum to exam #1, and R, +, juxtaposition, 0 is the usual vector space
structure (on the real number line). For v ∈ V define F (v) = ln(v).
Show that F : V → R is a linear mapping.
This really breaks down into two problems. First, to check that
F : V → R is a mapping, and second, to check that F is linear. The
first part is pretty easy since V = (0, ∞) and if v ∈ V (i.e. v is a
positive real number) then F (v) = ln(v) is a well-defined real number.
So F : V → R is a mapping.
The second part, to check that F is linear, involves applying the
definition of linear we stated above. This means we must verify that
the context of the term ‘linear’ is satisfied in our specific problem and
then work out restatements (in the notation of the problem) of the two
conditions (1) and (2) that we need to check in order to show that
F is linear. Once we have done that the step involving application of
the definition of linear is complete. But the problem itself will not be
complete until both condition (1) and condition (2) are verified.
We begin by interpreting or making specific in our problem the symbols in the definition that needed such interpretation.
• V in the definition is interpreted as (0, ∞) in the problem.
• + in the definition is interpreted in the problem as the mapping
(0, ∞) × (0, ∞) → (0, ∞) : (v1 , v2 ) 7→ v1 v2 (i.e. multiplication
of two positive real numbers).
• · in the definition is interpreted in the problem as the mapping
(0, ∞) × R → (0, ∞) : (v, a) 7→ va (i.e a positive real number
raised to a real number power).
• 0 in the definition is interpreted in the problem as the real
number 1.
• W in the definition is interpreted as R in the problem.
• +′ in the definition is interpreted in the problem as the mapping
R × R → R : (w1 , w2 ) 7→ w1 + w2 (i.e. addition of two real
numbers).
• ·′ in the definition is interpreted in the problem as the mapping
R × R → R : (w, a) 7→ wa (i.e a real number w multiplied by a
real number a). The word juxtaposition just indicates that we
do not write anything between w and a to denote the vector–
scalar product of one by the other.
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• 0′ in the definition is interpreted in the problem as the real
number 0.
• F in the definition is interpreted as the mapping v 7→ ln(v) in
the problem.
The context–check that V, +, ·, 0 and W, +′ , ·′ , 0′ are vector spaces and
F : V → W is a mapping (in the language of the definition) are satisfied
in the problem because in the addendum one checked that V, +, ·, 0 is a
vector space, and we know that R, +, juxtaposition, 0 is a vector space
and F : (0, ∞) → R : v 7→ ln(v) is a mapping. Condition (1) in the
definition:
For all v1 , v2 ∈ V we have that F (v1 + v2 ) = F (v1 ) + ′ F (v2 ),
becomes restated in the problem as follows:
For all v1 , v2 ∈ (0, ∞) we have that ln(v1 v2 ) = ln(v1 ) + ln(v2 ).
Likewise condition (2) in the definition:
For all v ∈ V and a ∈ R we have that F (v · a) = F (v) ·′ a,
becomes restated in the problem as follows:
For all v ∈ (0, ∞) and a ∈ R we have that ln(va ) = ln(v)a.
These two restatements are then immediate to verify, seeing as they
are well-known properties of the natural logarithm.