# Infinite rings and fields with positive characteristic

Familiar to us are infinite fields whose characteristic is equal to zero like $$\mathbb Z, \mathbb Q, \mathbb R$$ or the field of constructible numbers.

We’re also familiar with rings having infinite number of elements and zero for characteristic like:

• The rings of polynomials $$\mathbb Z[X], \mathbb Q[X], \mathbb R[X]$$.
• The rings of matrices $$\mathcal{M}_2(\mathbb R)$$.
• Or the ring of real continuous functions defined on $$\mathbb R$$.

We also know rings or fields like integers modulo $$n$$ (with $$n \ge 2$$) $$\mathbb Z_n$$ or the finite field $$\mathbb F_q$$ with $$q=p^r$$ elements where $$p$$ is a prime.

We provide below examples of infinite rings or fields with positive characteristic.

### Infinite rings with positive characteristic

Consider the ring $$\mathbb Z_n[X]$$ of polynomials in one variable $$X$$ with coefficients in $$\mathbb Z_n$$ for $$n \ge 2$$ integer. It is an infinite ring since $$\mathbb X^m \in \mathbb{Z}_n[X]$$ for all positive integers $$m$$, and $$X^r \neq X^s$$ for $$r \neq s$$. But the characteristic of $$\mathbb Z_n[X]$$ is clearly $$n$$.

Another example is based on product of rings. If $$I$$ is an index set and $$(R_i)_{i \in I}$$ a family of rings, one can define the product ring $$\displaystyle \prod_{i \in I} R_i$$. The operations are defined the natural way with $$(a_i)_{i \in I} + (b_i)_{i \in I} = (a_i+b_i)_{i \in I}$$ and $$(a_i)_{i \in I} \cdot (b_i)_{i \in I} = (a_i \cdot b_i)_{i \in I}$$. Fixing $$n \ge 2$$ integer and taking $$I = \mathbb N$$, $$R_i = \mathbb Z_n$$ for all $$i \in I$$ we get the ring $$\displaystyle R = \prod_{k \in \mathbb N} \mathbb Z_n$$. $$R$$ multiplicative identity is the sequence with all terms equal to $$1$$. The characteristic of $$R$$ is $$n$$ and $$R$$ is obviously infinite. Continue reading Infinite rings and fields with positive characteristic

# Playing with images and inverse images

We’re looking here to relational expressions involving image and inverse image.

We consider a function $$f : X \to Y$$ from the set $$X$$ to the set $$Y$$. If $$x$$ is a member of $$X$$, $$f(x)$$ is the image of $$x$$ under $$f$$.
The image of a subset $$A \subset X$$ under $$f$$ is the subset (of $$Y$$) $f(A)\stackrel{def}{=} \{f(x) : x \in A\}.$

The inverse image of a subset $$B \subset Y$$ is the subset of $$A$$ $f^{-1}(B)\stackrel{def}{=} \{x \in X : f(x) \in B\}.$ Important to understand is that here, $$f^{-1}$$ is not the inverse function of $$f$$.

We now look at relational expressions involving the image and the inverse image under $$f$$.

### Inverse images with unions and intersections

Following relations hold:
$\begin{array}{c} f^{-1}(B_1 \cup B_2) = f^{-1}(B_1) \cup f^{-1}(B_2)\\ f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2) \end{array}$ Let’s prove the first equality as an example. We have $$x \in f^{-1}(B_1 \cup B_2)$$ if and only if $$f(x) \in B_1 \cup B_2$$ if and only if $$f(x) \in B_1$$ or $$f(x) \in B_2$$ which means exactly $$x \in f^{-1}(B_1) \cup f^{-1}(B_2)$$. Continue reading Playing with images and inverse images

# Counterexamples around balls in metric spaces

Let’s play with balls in a metric space $$(M,d)$$. We denote by

• $$B_r(p) = \{x \in M : d(x,p) < r\}$$ the open ball.
• $$B_r[p] = \{x \in M : d(x,p) \le r\}$$ the closed ball.

### A ball of radius $$r$$ included in a ball of radius $$r^\prime < r$$

We take for $$M$$ the space $$\{0\} \cup [2, \infty)$$ equipped with the standard metric distance $$d(x,y)=\vert x – y \vert$$.

We have $$B_4(0) = \{0\} \cup [2, 4)$$ while $$B_3(2) = \{0\} \cup [2, 5)$$. Despite having a strictly smaller radius, the ball $$B_3(2)$$ strictly contains the ball $$B_4(0)$$.

The phenomenon cannot happen in a normed vector space $$(M, \Vert \cdot \Vert)$$. For the proof, take two open balls $$B_r(p),B_{r^\prime}(p^\prime) \subset M$$, $$0 < r^\prime < r$$ and suppose that $$p \in B_{r^\prime}(p^\prime)$$. If $$p=p^\prime$$ and $$q \in B_{r^\prime}(p^\prime) \setminus \{p^\prime\}$$ then $$p + \frac{\frac{r+r^\prime}{2} }{\Vert p q \Vert} p q \in B_r(p) \setminus B_{r^\prime}(p^\prime)$$. And if $$p \neq p^\prime$$, $$p \in B_{r^\prime}(p^\prime)$$ then $$p^\prime + \frac{\frac{r+r^\prime}{2} }{\Vert p^\prime p \Vert} p^\prime p \in B_r(p) \setminus B_{r^\prime}(p^\prime)$$.

### An open ball $$B_r(p)$$ whose closure is not equal to the closed ball $$B_r[p]$$

Here we take for $$M$$ a subspace of $$\mathbb R^2$$ which is the union of the origin $$\{0\}$$ with the unit circle $$S^1$$. For the distance, we use the Euclidean norm.
The open unit ball centered at the origin $$B_1(0)$$ is reduced to the origin: $$B_1(0) = \{0\}$$. Its closure $$\overline{B_1(0)}$$ is itself. However the closed ball $$B_1[0]$$ is the all space $$\{0\} \cup S^1$$.

Again one can prove that for a normed vector space this cannot happen. The closure of an open ball is the closed ball for a normed vector space.

# Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If $$K$$ is a compact topological space, and $$(f_n)_{n \in \mathbb N}$$ is a monotonically decreasing sequence (meaning $$f_{n+1}(x) \le f_n(x)$$ for all $$n \in \mathbb N$$ and $$x \in K$$) of continuous real-valued functions on $$K$$ which converges pointwise to a continuous function $$f$$, then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

### Cases if $$K$$ is not compact

We take $$K=(0,1)$$, which is not closed equipped with the common distance. The sequence $$f_n(x)=x^n$$ of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all $$n \in \mathbb N$$ $\sup\limits_{x \in (0,1)} x^n = 1$

The set $$K=\mathbb R$$ is closed but unbounded, hence also not compact. The sequence defined by $f_n(x)=\begin{cases} 0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}$ is continuous and monotonically decreasing. It converges to $$0$$. However, the convergence is not uniform as for all $$n \in \mathbb N$$: $$\sup\{f_n(x) : x \in \mathbb R\} =1$$. Continue reading Counterexamples around Dini’s theorem