olivierd – Precisely

Coprimality with a product of integers

The theorem

If $n > 0$ and $a, b_1, \ldots b_n \in \Z$ are such that $a$ is separately coprime with each $b_i$ , then $a$ is coprime with their product $b_1 b_2 \ldots b_n$ .

The proof

Proof: Through Bézout. $a$ being coprime with each $b_i$ , we can write, for each $i = 1, \ldots n$ :

$p_i a + q_i b_i = 1$

If we take the product of each of these expressions, we get:

$\prod_{i = 1}^n (p_i a + q_i b_i) = 1$

Among the $2^n$ terms of this product, all but one contain $a$ as a factor. The one that doesn’t have $a$ as a factor is $q_1 b_1 q_2 b_2 \ldots q_n b_n$ . Hence we obtain an expression of the form:

$(\ldots) a + (q_1 \ldots q_n) (b_1 \ldots b_n) = 1$

Hence $a$ and $b_1 b_2 \ldots b_n$ are coprime. $\blacksquare$

In the case of polynomials

The same result holds with polynomials: if $P$ and $Q_1, \ldots Q_n$ are polynomials over a field $\K$ , such that $P$ is coprime with each $Q_i$ , then $P$ is coprime with their product $Q_1 Q_2 \ldots Q_n$ .

The proof is identical.

Degree and Euclidean division of polynomials

For the definition and characterization of the polynomials over a field $\K$ , see Polynomials over a field.

Definition of the degree of a polynomial

The degree of a polynomial P, written $\deg P$ , is an element of $\{-\infty\} \cup \N$ . The symbol $-\infty$ is taken to have the usual properties of order and of addition relative to natural numbers.

The family $(X^0, X^1, X^2, \ldots)$ is a $\K$ -vector space basis of $\K[X]$ . For any $P \in \K[X]$ , there exists an unique $(a_0, a_1, \ldots) \in \K^\N$ , only finitely many of which are nonzero, such that $P = \sum_n a_n X^n$ . If all the $a_n$ are zero, that is, if $P$ is the zero element of $\K[X]$ , then $\deg P$ is defined as $-\infty$ . If not all the $a_n$ are zero, since only finitely many are nonzero, there is a greatest value of $n$ such that $a_n \neq 0_\K$ . Then $\deg P$ is defined as this greatest value.

Elementary properties of the degree

The following are easily checked to hold for any polynomials $P$ and $Q$ (even when one or both are zero):

$\deg (P + Q) \le \sup(\deg P, \deg Q)$

if $\deg P \neq \deg Q$ , then $\deg (P + Q) = \sup(\deg P, \deg Q)$

$\deg PQ = \deg P + \deg Q$

Euclidean division

The notion of the degree of a polynomial allows us to formulate the following property of Euclidean division:

For all polynomials $A$ and $B$ , with $B$ nonzero, there exists exactly one pair of polynomials $(Q, R)$ such that $A = Q B + R$ with $\deg R < \deg B$ .

For the proof, we fix the value of nonzero polynomial $B$ , and recurse over the degree of $A$ .

More specifically, $B$ being given, we consider the following proposition dependent on $n \in \{-\infty\} \cup \N$ :

$\mathcal P(n)$ iff for all polynomials $A$ such that $\deg A \le n$ , there exist polynomials $Q, R$ , with $\deg R < \deg B$ , such that $A = Q B + R$ .

$\mathcal P(-\infty)$ is trivially true (for $\deg A \le -\infty$ means that $A$ is zero, and $Q = 0, R = 0$ satisfies $A = Q B + R$ with $\deg R < \deg B$ ).

Let us suppose $\mathcal P(n)$ true for some $n \in \{-\infty\} \cup \N$ , and consider a polynomial $A$ the degree of which is less or equal to the successor of $n$ , that is to $0$ if $n = -\infty$ , or $n+1$ otherwise.

If the degree of $A$ is less or equal to $n$ , then $\mathcal P(n)$ applies and there exist the wanted $Q$ and $R$ .

If the degree of $A$ is less than that of $B$ , then we can directly write $A = Q B + R$ with $Q = 0$ and $R = A$ .

Remains the case in which $\deg A$ is the successor of $n$ .

Polynomials over a field

In this context, we consider a commutative field (simply: field) $\K$ . The polynomials will be constructed as a certain associative unital algebra over $\K$ (“ $\K$ -aua”), together with a distinguished element called the formal variable.

$\K$ itself, as a vector space over itself and the usual multiplication in $\K$ as the third law, is a $\K$ -aua,

Morphisms between two $\K$ -auas are linear mappings that conserve the third law and map the unit of the first algebra onto the unit of the second one.

All that will be said of the polynomials over $\K$ will be based on the universal property given below. The polynomials will thus be defined uniquely up to an isomorphism.

Definition

The polynomials over $\K$ is any $\K$ -aua $A$ together with a distinguished element $X$ of $A$ such that the following universal property holds:

For any $\K$ -aua $A'$ and any $a \in A$ , there exists exactly one morphism $\phi: A \to A'$ such that $\phi(X) = a$ .

Unicity up to an isomorphism

Let $(A_1, X_1)$ and $(A_2, X_2)$ be two sets of polynomials over $\K$ . The universal property implies the existence of a morphism $\phi_1: A_1 \to A_2$ such that $\phi(X_1) = X_2$ , and of a morphism $\phi_2: A_2 \to A_1$ such that $\phi(X_2) = X_1$ .

We then have $(\phi_2 \circ \phi_1)(X_1) = X_1$ . Hence $\phi_2 \circ \phi_1$ is a morphism $A_1 to A_1$ that maps $X_1$ to $X_1$ . Now $\Id_{A_1}$ is another such morphism. By virtue of the universal property of $(A_1, X_1)$ , it follows that $\phi_2 \circ \phi_1 = \Id_{A_1}$ , that is, is the $\K$ -aua category identity on $A_1$ .

The same reasoning shows that $\phi_1 \circ \phi_2$ is the $\K$ -aua category identity on $A_2$ .

Hence $\phi_1$ is an isomorphism $A_1 \to A_2$ that maps the formal variable of $A_1$ to that of $A_2$ .

Characterization

Let $A$ be an associative unital algebra over $\K$ , and $X$ an element of $A$ . Then $(A, X)$ represents the polynomials over $\K$ if and only if the family $(X^n)_{n \in \N}$ , with $X^0 = 1_A$ and $X^{n+1} = X X^n$ , is a family-base of the $\K$ -vector space $A$ .

Let us first suppose that the latter condition holds on $(A, X)$ , and prove the universal property.

Let $A'$ be a $\K$ -aua and $a'$ an element of $A'$ .

Let us suppose that $\phi$ is a morphism $A \to A'$ such that $\phi(X) = a'$ . Then for all $n \in \N$ , we have $\phi(X^n) = a'^n$ . Since $\phi$ as a $\K$ -aua morphism, it is in particular a linear mapping. Since a linear mapping is defined by the image of a basis, and $(X^n)_{n \in \N}$ is a basis, $\phi$ necessarily is the unique linear mapping $A \to A'$ such that $\forall n \in \N, \phi(X^n) = a'^n$ . Now it is easy to check that this linear mapping does preserve the multiplication in $A$ , and is hence a $\K$ -aua morphism.

Thus the universal property is proven for $(A, X)$ .

Let us now suppose the universal property for $(A, X)$ , and show that $(X^n)_{n \in \N}$ is a family-base of the $\K$ -vector space $A$ .

Let us first show that $(X^n)_{n \in \N}$ is linearly independent.

Let $(A', \alpha)$ be a free $\K$ -vector space on set $\N$ . This entails that $(\alpha(n))_{n \in \N}$ is a family-base of $A'$ . To make $A'$ into an associative unital algebra, we must define a multiplication law that is bilinear, and also associative.

A bilinear law can be defined by the images of all the ordered pairs of base elements. Hence we may define the multiplication in $A'$ by:

$\forall n_1, n_2 \in \N, \alpha(n_1) \cdot \alpha(n_2) = \alpha(n_1 + n_2)$

We then have $(\alpha(n_1) \cdot \alpha(n_2)) \cdot \alpha(n_3) =$ $\alpha(n_1 + n_2) \cdot \alpha(n_3) =$ $\alpha(n_1 + n_2 + n_3) =$ $\alpha(n_1) \cdot \alpha(n_2 + n_3) =$ $\alpha(n_1) \cdot (\alpha(n_2) \cdot \alpha(n_3))$ .

Hence the multiplication is associative with regard to the elements of the base; it is easy to check that this follow through to arbitrary elements, which makes $A'$ an associative algebra.

$\alpha(0)$ is a unit; hence $A'$ is an associative unital algebra.

Applying the universal property of $(A, X)$ , we find that there exists exactly one $\K$ -aua morphism $\phi: A \to A'$ such that $\phi(X) = \alpha(1)$ .

We then have $\phi(X^0) = \phi(1_A) = 1_{A'} =$ $\alpha(0)$ ; $\phi(X^1) =$ $\phi(X) = \alpha(1)$ ; and, for any $n > 1$ , $\phi(X^n) =$ $\phi(X \cdot \ldots \cdot X) =$ $\phi(X) \cdot \ldots \cdot \phi(X) =$ $\alpha(1) \cdot \ldots \cdot \alpha(1) =$ $\alpha(1 + \ldots + 1) = \alpha(n)$ . Hence for all $n \in \N$ , $\phi(X^n) = \alpha(n)$ .

Let us consider a linear combination of the family $(X^n)_{n \in \N}$ that vanishes; that is, a finite sequence of values of $\K$ , $(a_0, a_1, \ldots a_n)$ , such that $\sum_{i = 0}^n a_i X^i = 0_A$ . Taking the image of this linear combination by $\phi$ , we obtain $\sum_{i = 0}^n a_i \alpha(i) = 0_{A'}$ . Since the $\alpha(i)$ form a basis, all the $a_i$ must be zero.

Thus the family $(X^n)_{n \in \N}$ is linearly independent.

Lastly, we must show that this family generates the $\K$ -vector space $A$ .

Let $B$ be the sub-aua of $A$ generated by $\{X\}$ .

There exists a unique aua morphism $\phi: A \to B$ such that $\phi(X) = X$ .

Let then $\phi'$ be $\phi$ , but with all of $A$ as codomain; in other words, $\phi': A \to A, x \mapsto \phi(x)$ . It is clear that $\phi'$ too is an aua morphism, and it is such that $\phi'(X) = X$ .

But there is exactly one aua morphism $A \to A$ that maps $X$ to $X$ ; and $\Id_A$ is such.

Hence $\phi' = \Id_A$ .

The image of $\phi'$ is that of $\phi$ , and is a subset of $B$ . But the image of $\Id_A$ is $A$ itself. Hence $B = A$ .

Thus the sub-aua of $A$ generated by $\{X\}$ is $A$ itself. This sub-aua is the set of linear combinations of all the powers of $X$ ; hence the sub-vector space generated by $X^0, X^1, X^2, \ldots$ is $A$ .

Thus $(X^n)_{n \in \N}$ is a family-base of the $\K$ -vector space $A$ .

The composition of polynomials

Definition

Let $\K$ be a commutative field. The polynomials $\K[X]$ form a commutative unital algebra over $\K$ containing an element $X$ such that the following universal property holds:

For any unital algebra $A$ (commutative or otherwise) and any element $a \in A$ , there exists exactly one unital algebra morphism $\phi: \K[X] \to A$ such that $\phi(X) = a$ .

If $P$ and $Q$ are elements of $\K[X]$ , let $\gamma_Q: \K[X] \to \K[X]$ be the unique morphism such that $\gamma_Q(X) = Q$ . Then the composition $P(Q)$ of $P$ and $Q$ is defined by $P(Q) = \gamma_Q(P)$ .

Effect of composition of polynomials on the corresponding functions

In terms of polynomial functions, the composition of polynomials is the equivalent to the composition (via “ $\circ$ ”) of the associated functions.

Let us first define the function that is associated with a polynomial.

The set $\K^\K$ of functions $\K \to \K$ is made into a commutative and unital associative algebra by the usual operations of summing ( $(f + g)(x) = f(x) + g(x)$ , multiplication by a scalar ( $(af)(x) = a(f(x))$ and internal multiplication ( $(fg)(x) = f(x) g(x)$ ).

Hence, following the universal property of the algebra of polynomials, there exists a unique morphism $\phi: \K[X] \to \K$ such that $\phi(X) = \Id_\K$ .

By definition, the function $p$ associated with a polynomial $P$ is simply $p = \phi(P)$ .

We wish to show that if $P$ and $Q$ are polynomials and $p$ and $q$ the respective associated functions, the function associated with $P(Q)$ is $p \circ q$ .

Let $Q$ be a polynomial and $q = \phi(Q)$ . Let us consider the mapping $\alpha: \K[X] \to \K^\K, P \mapsto \phi(P) \circ q$ .

It is easy to check that $\alpha$ is a morphism.

Furthermore, $\alpha(X) = \phi(X) \circ q = \Id_\K \circ q = q$ .

On the other hand, $\phi \circ \gamma_Q$ is also a morphism from $\K[X]$ to $\K^\K$ , and $(\phi \circ \gamma_Q)(X) = \phi(\gamma_Q(X)) = \phi(Q) = q$ .

Thus both $\alpha$ and $\phi \circ \gamma_Q$ are morphisms $\K[X] \to \K^\K$ and they agree on the value $X$ . The universal property of polynomials entails that they are equal.

Hence for any $P \in \K[X]$ , $(\phi \circ \gamma_Q)(P) = \alpha(P)$ .

That is, $\phi(P(Q)) = \phi(P) \circ \phi(Q)$ .

Effect on the evaluation of a polynomial

If $P \in \K[X]$ and $a \in \K$ , the evaluation of $P$ at $a$ is, by definition, the value at $P$ of the unique morphism $\epsilon_a: \K[X] \to \K$ such that $\epsilon_a(X) = a$ .

Given polynomials $P$ and $Q$ and $a \in \K$ , we wish to evaluate $P(Q)$ at $a$ , and, specifically, show that $\epsilon_a(P(Q)) = \epsilon_{\epsilon_a(Q)}(P)$ .

Since $P(Q) = \gamma_Q(P)$ , we have $\epsilon_a(P(Q)) = (\epsilon_a \circ \gamma_Q)(P)$ .

Both $\epsilon_a \circ \gamma_Q$ and $\epsilon_{\epsilon_a(Q)}$ are morphisms $\K[X] \to \K$ . Both evaluate at $X$ to $\epsilon_a(Q)$ . Hence they are equal:

$\epsilon_a \circ \gamma_Q = \epsilon_{\epsilon_a(Q)}$

which entails $(\epsilon_a \circ \gamma_Q)(P) = \epsilon_{\epsilon_a(Q)}(P)$

hence: $\epsilon_a(P(Q)) = \epsilon_{\epsilon_a(Q)}(P)$ , as announced.

In particular: if $P \in \K[X]$ and $a, b \in \K$ , then $\epsilon_a(P(X + b)) = \epsilon_{\epsilon_a(X + b)}(P) = \epsilon_{\epsilon_a(X) + \epsilon_a(b)}(P) = \epsilon_{a + b}(P)$ .

Associativity

If $P$ , $Q$ and $R$ are polynomials in $\K[X]$ , we wish to show that:

$P(Q(R)) = (P(Q))(R)$

Using the definition of composition, this translates into:

$\gamma_{\gamma_R(Q)}(P) = \gamma_R(\gamma_Q(P))$

That is:

$\gamma_{\gamma_R(Q)}(P) = (\gamma_R \circ \gamma_Q)(P)$

We need to show that for any $Q$ and $R$ in $\K[X]$ , the two $\K[X] \to \K[X]$ mappings $\gamma_{\gamma_R(Q)}$ and $\gamma_R \circ \gamma_Q$ are identical.

Both are endomorphisms of $\K[X]$ .

We have:

$\gamma_{\gamma_R(Q)}(X) = \gamma_R(Q)$ by definition of $\gamma_{\gamma_R(Q)}$ .
$(\gamma_R \circ \gamma_Q)(X) = \gamma_R(\gamma_Q)(X)) = \gamma_R(Q)$ by definition of $\gamma_Q)$ .

Hence both morphisms agree on the image of $X$ . By virtue of the universal property of polynomials, they are equal.

Hence for any $P \in \K[X]$ , we indeed have $\gamma_{\gamma_R(Q)}(P) = (\gamma_R \circ \gamma_Q)(P)$ , that is, $P(Q(R)) = (P(Q))(R)$ .

Effect of composition on the evaluation of polynomials

todo

Derivation of the composition of polynomials

todo

Equivalent alternative definitions concerning filters

Definition of a filter

Let $E$ be a set.

Traditional definition of a filter on $E$ : A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff $\emptyset \notin E$ ; and $\forall A \in \mathcal F, \forall B \subseteq E, (B \supseteq A \implies B \in \mathcal F)$ ; and $\forall A, B \in \mathcal F, A \cap B \in \mathcal F$ .

Alternative definition: A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff $\emptyset \notin E$ ; and $\forall A, B \subseteq E, ((A, B) \in \mathcal F^2 \iff A \cap B \in \mathcal F)$ .

The two definitions are equivalent.

Definition of the image of a filter

Let $E$ and $F$ be sets, $\mathcal F$ a filter on $E$ and $f: E \to F$ .

Traditional definition: The filter on $F$ image of $\mathcal F$ by $f$ is $(f^{\rightarrow \rightarrow}(\mathcal F))^{\uparrow F}$ .

Alternative definition: The filter on $F$ image of $\mathcal F$ by $f$ is $\{\ B \subseteq F\ |\ f^\leftarrow(B) \in \mathcal F\ \}$ .

Again, the two definitions are equivalent.

Margin between a compact subspace and an including open subspace

In Chapter 31 (“Compact-Open Topology”), the fact that, on the set of the continuous mappings $\R \to \R$ the second topology (the uniform convergence topology) is finer than the first (the compact-open topology) is said to be “clear” but actually needs some non-trivial proof. Part of this is the lemma that I state and prove below.

For any $x \in E$ and real $r > 0$ , let $\dot B(x, r)$ be the open ball centered on $x$ and with radius $r$ . We know that an open ball is an open subset.

The lemma says:

Let $(E, d)$ be a metric space, $O$ an open subspace of $E$ and $C \subseteq O$ a compact subspace of $E$ . Then there exists $r > 0$ such that for all $x$ in $C$ , $\dot B(x, r) \subseteq O$ .

Proof:

For each $x \in C$ , since $x \in O$ , there exists an $r_x > 0$ such that $\dot B(x, r_x) \subseteq O$ .

The collection of all $\dot B(x, \frac 1 2 {r_x})$ with $x \in C$ is an open cover of $C$ . Hence there exists a finite $J \subseteq C$ such that $\{\ \dot B(x, \frac 1 2 {r_x})\ \}_{x \in J}$ covers $C$ .

Since $J$ is finite, there exists a real $r > 0$ such that for all $x \in J$ , $r \le \frac 1 2 {r_x}$ .

Let $x$ be an eventual point of $C$ .

Since $\{\ \dot B(z, \frac 1 2 {r_z})\ \}_{z \in J}$ covers $C$ , there exists $z \in C$ such that $x \in \dot B(z, \frac 1 2 {r_z})$ , that is, $d(z, x) < \frac 1 2 {r_z}$ .

Let $y$ be an eventual point of $\dot B(x, r)$ .

We have $d(z, x) < \frac 1 2 {r_z}$ and $d(x, y) < r \le \frac 1 2 {r_z}$ , which leads to $d(z, y) < r_z$ , that is, $y \in \dot B(z, r_z)$ . But $z$ is in $C$ and $r_z$ was taken such that $\dot B(z, r_z) \subseteq O$ . Hence $y \in O$ .

Hence $\dot B(x, r) \subseteq O$ .

We thus have $r > 0$ such that $\forall x \in C,\ \dot B(x, r) \subseteq O$ . $\blacksquare$

Open tubes

In Chapter 31, “The Compact-Open Topology”, it is asserted that “clearly” the third topology on $\Mor(X, Y)$ (with $X$ and $Y$ the real line) is finer than the second. The issue is more fully discussed here.

In this post I discuss the concept of open tubes, particular subsets of the Cartesian product $X \times Y$ with $X$ a set and $Y$ a metric space.

Definition

Let $X$ be a set and $Y$ a metric space, with distance $d$ .

For any $r > 0$ and any $a \in Y$ , we note $\dot B(a, r)$ the open ball of $Y$ of radius $r$ and centred on $a$ , that is set of all $x \in Y$ such that $d(a, x) < r$ . Open balls are notoriously open subsets.

Let $\phi$ be any mapping $X \to Y$ , and $r$ a real $> 0$ .

Then the open tube $\dot T(\phi, r)$ of radius $r$ and centred on $\phi$ is the following subset of $X \times Y$ :

$\dot T(\phi, r) = \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}$

Equivalently:

$\dot T(\phi, r) = \bigcup_{x \in X} \{x\} \times \dot B(\phi(x), r)$

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to $X$ , so we don’t have any particular one on $X \times Y$ and cannot speak of $\dot T(\phi, r)$ being open or not.

However, if $X$ is a topological space and $\phi$ is continuous $X \to Y$ , then $\dot T(\phi, r)$ is open, as I will show.

An open tube centred on a continuous mapping is open

If $X$ is a topological space and $\phi$ is a continuous mapping $X \to Y$ , then any open tube centred on $\phi$ is an open subset of $X \times Y$ .

Proof:

I will show that in the above circumstances $\dot T(\phi, r)$ is a neighbourhood of each of its points.

Let $(x_0, y_0)$ be an eventual point of $\dot T(\phi, r)$ .

Then $d(\phi(x_0), y_0) < r$ .

Lettuce define $s = \frac 1 2 (r - d(\phi(x_0), y_0))$ . We have $s > 0$ .

Let $D = \phi^\leftarrow(\dot B(\phi(x_0), s))$ . Since $\phi$ is continuous and $\dot B(\phi(x_0), s)$ is an open subset of $Y$ , $D$ is an open subset of $X$ . Since $\phi(x_0) \in \dot B(\phi(x_0), s)$ , $x_0 \in D$ .

Let $O = D \times \dot B(y_0, s)$ , open subset of $X \times Y$ . Clearly, $(x_0, y_0) \in O$ , so $O$ is a neighbourhood of $(x_0, y_0)$ .

Let $(x, y)$ be an eventual point of $O$ .

$x \in D$ , hence $\phi(x) \in \dot B(\phi(x_0), s)$ , that is $d(\phi(x), \phi(x_0)) < s$ .

$y \in \dot B(y_0, s)$ , hence $d(y_0, y) < s$ .

$d(\phi(x), y) \le d(\phi(x), \phi(x_0)) + d(\phi(x_0), y_0) + d(y_0, y)$

Since $d(\phi(x), \phi(x_0)) < s$ and $d(y_0, y) < s$ , this leads to:

$d(\phi(x), y) < s + d(\phi(x_0), y_0) + s$

With:

$s + d(\phi(x_0), y_0) + s = 2 s + d(\phi(x_0), y_0)$
$\ = 2 (\frac 1 2 (r - d(\phi(x_0), y_0))) + d(\phi(x_0), y_0)$
$\ = r$

Thus we have: $d(\phi(x), y) < r$ .

Which implies: $(x, y) \in \dot T(\phi, r)$ .

Hence $O \subseteq \dot T(\phi, r)$ .

Thus we have found a neighbourhood $O$ of $(x_0, y_0)$ that is a subset of $\dot T(\phi, r)$ , which implies that $\dot T(\phi, r)$ itself is a neighbourhood of $(x_0, y_0)$ .

Thus $\dot T(\phi, r)$ is an open subset of $X \times Y$ . $\blacksquare$

Uniform convergence topology and the open set topology

Chapter 31, “The Compact-Open Topology”, describes the compact-open topology, and goes on to compare it with two other topologies on $\Mor(X, Y)$ in the case where $X = Y = \R$ . It asserts that “clearly” the second of these three is finer than the first, and the third finer than the second.

This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that $X$ is a topological space and $Y$ a metric space with distance $d$ .

I define $E = \Mor(X, Y)$ the set of continuous mappings from $X$ to $Y$ .

I note $\dot B(y, r)$ the open ball in $Y$ centred on $y \in Y$ and with radius $r > 0$ . It is the set of all $y' \in Y$ such that $d(y, y') < r$ .

Furthermore, for any $H$ subset of $X \times Y$ , I note $\kappa(H)$ the set of all continuous mappings $\phi: X \to Y$ the graph of which is a subset of $H$ ; in other words, $\kappa(H) = \{\ \phi \in E\ |\ \forall x \in X, (x, \phi(x)) \in H\ \}$ .

The uniform convergence topology on $E$ is such that a subset $A$ of $E$ is open if and only if for any $\phi \in A$ , there exists $r > 0$ such that for any $\psi \in E$ , if $\forall x \in X, \psi(x) \in \dot B(\phi(x), r)$ , then $\psi \in A$ . We can express this condition in terms of subsets of $X \times Y$ that I have named open tubes: for any mapping (not necessarily continuous) $\phi: X \to Y$ and for any $r > 0$ , the open tube $\dot T(\phi, r)$ is $\{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}$ . Then $A \subseteq E$ is open for the uniform convergence topology if and only if for any $\phi \in A$ , there exists $r > 0$ such that $\kappa(\dot T(\phi, r)) \subseteq A$ .

The third topology ‑ the “open set” topology ‑ is the topology on $E$ generated by all the $\kappa(O)$ for all $O$ open subsets of $X \times Y$ . (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let $A$ be an open set in the uniform convergence topology.

Let $\phi$ be an eventual member of $A$ .

There exists an $r > 0$ such that $\kappa(\dot T(\phi, r)) \subseteq A$ .

I prove here that, $\phi$ being continuous, $\dot T(\phi, r)$ is an open subset of $X \times Y$ . It includes the graph of $\phi$ . Hence $\kappa(\dot T(\phi, r))$ contains $\phi$ , and is by definition an open set for the open set topology on $E$ . Since it is a subset of $A$ , $A$ is a neighbourhood of $\phi$ for the open set topology.

Hence for any $\phi \in A$ , $A$ is a neighbourhood of $\phi$ for the open set topology. This implies that $A$ is itself an open set in the open set topology. $\blacksquare$

Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let $A$ and $B$ be connected subsets of topological space $X$ such that $A$ intersects $\Cl(B)$ . Prove that $A \cup B$ is connected. Find an example to show that it is not enough to assume that $\Cl(A)$ intersects $\Cl(B)$ .

Let us assume that $V$ and $W$ are open subsets of $X$ such that $V \cap W \cap (A \cup B)$ is empty and that $A \cup B \subseteq V \cup W$ . To show that $A \cup B$ is connected, we need to show that either $V$ or $W$ does not intersect $A \cup B$ .

Since $A$ is connected, and $V \cap W \cap A$ is empty and $A \subseteq V \cup W$ , necessarily $A$ is entirely in $V$ or in $W$ . Similarly, $B$ is entirely in $V$ or entirely in $W$ .

Let us suppose that they are not both in $V$ or both in $W$ ; for instance, that $A \subseteq V$ and $B \subseteq W$ .

If $a \in A$ , then $a \in V$ with $V$ open in $X$ . Since $V$ does not intersect $B$ , $a$ cannot be in $\Cl(B)$ . Hence $A$ cannot intersect $\Cl(B)$ . This contradicts our assumption.

Hence it is impossible that $A \subseteq V$ and $B \subseteq W$ . Similarly, it is impossible that $A \subseteq W$ and $B \subseteq V$ . Thus, $A$ and $B$ are either both in $V$ or both in $W$ , which implies that either $V$ or $W$ does not intersect $A \cup B$ . $\blacksquare$

If we only assume that $\Cl(A)$ and $Cl(B)$ intersect, it does not always follow that $A \cup B$ is connected. For instance, let $X$ be the topological plane and $A$ and $B$ two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of $A$ and $B$ is then not connected.

Union of transitively overlapping connected spaces

Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let $A_\lambda$ ( $\lambda$ in $\Lambda \neq \emptyset$ ) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by $A_\lambda \approx A_{\lambda'}$ if $A_\lambda \cap A_{\lambda'} \neq \emptyset$ have just one equivalence class. Then $\bigcup_\lambda A_\lambda$ is connected.

For clarity, rather than a family $(A_\lambda)_{\lambda \in \Lambda}$ I will consider a set $\mathcal A$ of connected subsets of $X$ . The condition is that the equivalence relation in $\mathcal A$ generated by the above relation $\approx$ has only one class.

Let $K = \bigcup_{A \in \mathcal A} A$ .

Let $V$ and $W$ be eventual open subsets of $X$ such that $K \subseteq V \cup W$ and $V \cap W \cap K = \emptyset$ . To show that $K$ is connected, we must show that either $V \cap K$ or $W \cap K$ is empty.

First, let $A$ be an eventual element of $\mathcal A$ . Since $A \subseteq K$ , the two subsets of $A$ , $A \cap V$ and $A \cap W$ are disjoint, and their union is $A$ . Furthermore, they are open in $A$ . Since $A$ is connected, this implies that one or the other is empty. If $A \cap V$ is empty, then $A \subseteq W$ ; if instead $A \cap W$ is empty, then $A \subseteq V$ . Hence in all cases an element of $\mathcal A$ is entirely in $V$ or entirely in $W$ .

Let us consider the relation $\mathcal R$ on $\mathcal A$ defined by $A_1 \mathcal R A_2 \iff (A_1 \subseteq V \wedge A_2 \subseteq V) \vee (A_1 \subseteq W \wedge A_2 \subseteq W)$ ; in other words, $A_1 \mathcal R A_2$ if and only if both are subsets of $V$ , or both are subsets of $W$ . Clearly, this is an equivalence relation.

Furthermore, if $A_1 \approx A_2$ , that is, $A_1 \cap A_2 \neq \emptyset$ , we cannot have $A_1 \subseteq V$ and $A_2 \subseteq W$ , for $V$ and $W$ are disjoint. Hence $A_1$ and $A_2$ are either both in $V$ or both in $W$ ; that is, $A_1 \mathcal R A_2$ . Hence $\mathcal R$ is an equivalence relation greater than the equivalence relation generated by $\approx$ .

If $\mathcal R$ had two or more classes, there would be some element of $\mathcal A$ in one class and another in another; these two elements would not be equivalent following $\mathcal R$ . Since $\mathcal R$ is greater than the equivalence relation generated by $\approx$ , they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence $\mathcal R$ too can have only one class, which means that for all $A_1$ and $A_2$ in $\mathcal A$ , either both are in $V$ or both are in $W$ . But this implies that all are in $V$ or all are in $W$ , which in turn implies that their union $K$ is a subset of either $V$ or $W$ , hence that either $V \cap K$ or $W \cap K$ is empty. $\blacksquare$