Question

Give an example of a countable collection of algebraic sets whose union is not algebraic.

Short answer

Question: Provide an example of a countable collection of algebraic sets whose union is not algebraic. Answer: The countable collection of algebraic sets {A_n} for n = 1,2,3, ... satisfies the given condition, where A_n = {(x, y) ∈ R^2 | y = nx^2 } and their union is not algebraic.

Step-by-step solution

Recall Examples of Algebraic Sets

Algebraic sets are defined in terms of polynomial functions and their common zeros. For example, the zero sets of the polynomial functions like f(x) = x and g(x) = x^2 - 2 are algebraic sets. We now focus on finding a countable collection of algebraic sets whose union is not algebraic.

Define a countable collection of algebraic sets

Let's consider the countable family of algebraic sets {A_n}, where for each positive integer n, A_n is defined as A_n = {(x, y) ∈ R^2 | y = nx^2 }. These sets are algebraic since each A_n is the zero set of the polynomial function f(x, y) = y - nx^2.

Show the union of these algebraic sets is not algebraic

The union of these algebraic sets is a subset of R^2 defined as B = {(x, y) | y/n = x^2 for some positive integer n}. Assume, for the sake of contradiction, that this set B is algebraic. Then there exists a polynomial function g(x, y) whose zero set is B. Since B contains infinitely many parabolas y = nx^2 for n = 1,2,3, ... , the polynomial g(x, y) should have infinitely many factors, one for each of those parabolas. However, polynomials have a finite number of factors, which creates a contradiction. Therefore, the union B cannot be an algebraic set. So, the countable collection of algebraic sets {A_n} for n = 1,2,3, ... satisfies the given condition, where A_n = {(x, y) ∈ R^2 | y = nx^2 } and their union is not algebraic.

Key concepts

Polynomial Functions

Imagine a tool that allows us to construct curves and shapes with just a few mathematical expressions; this is where polynomial functions step in. A polynomial function is an expression consisting of variables, coefficients, and non-negative integer exponents. It looks something like this:
\( f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x^1 + a_0 \), where the coefficients \( a_n, a_{n-1}, \ldots, a_0 \) are real numbers, and \( n \) is a non-negative integer called the degree of the polynomial.

The beauty of polynomials lies in their simplicity and the diverse patterns they can generate, such as straight lines, parabolas, or more complex curves. They're like the building blocks for algebraic sets, creating the 'zero set' when the polynomial is equal to zero. This happens at specific points for, say, a line, or along a curve for a quadratic equation. In our textbook example, the zero set of \( f(x) = x \) is simply the origin (0,0), while the zero set of \( g(x) = x^2 - 2 \) includes points where the parabola intersects the x-axis.

How Do They Relate to Algebraic Sets?

Think of a child's dot-to-dot puzzle; polynomial functions give us the dots (zero sets), and our pencil line connecting them forms algebraic sets. Algebraic sets are collections of all the points that satisfy one or more polynomial equations, essentially highlighting the shapes these polynomials predict. The simple idea of setting a polynomial equal to zero gives rise to intricate geometric figures on coordinate planes.

Countable Union of Sets

Imagine you have an infinite pile of coins, and you can count them one by one. Even though it might take forever, it’s theoretically possible to count them with no coin being skipped. This is what mathematicians mean by a countable set. It's infinite but can be put into a one-to-one correspondence with natural numbers (1, 2, 3, ...).

Now, imagine you have several piles of coins, say a separate pile for each day of the year. Each pile is countable, and the number of piles is also countable (365 for a year, or 366 for a leap year). The union of all these piles is still countable - you could label each coin with a day and a number from its pile. This idea is mirrored in the concept of countable unions in set theory.
The countable union of sets takes individual countable sets and combines them into one larger set. It’s like merging multiple playlists into one mega playlist; each song (or element) is still countable within the larger list. Mathematically, it's expressed as \( \bigcup_{n=1}^{\infty} A_n \) for a series of sets \( A_1, A_2, ..., A_n \). Going back to our exercise, each algebraic set \( A_n \) is countable because it aligns with points on a parabola, and the union of these sets is countable overall. However, as the solution reveals, just because something is countable doesn't mean it will neatly fit into an algebraic set.

Zero Set of a Polynomial

Time to put on our detective hats and find out where a polynomial function equals zero. The zero set of a polynomial is like the 'whodunit' of algebra; it's a complete list of all the suspects (points) where the polynomial gets caught with a value of zero.

If a polynomial were a burglar, the zero set would be where it left fingerprints. Mathematically, for a polynomial \( f(x) \), the zero set is \( \{ x | f(x) = 0 \} \). For example, the zero set of \( f(x) = x^2 - 4 \) would be the points where \( x^2 - 4 \) equals zero, or where \( x = \pm 2 \).
In the context of our textbook exercise, the zero set for each polynomial \( f(x, y) = y - nx^2 \) is a parabola. All these parabolas together would look like a blueprint for a curvy mountain range.

But What Happens When We Take Their Union?

Here's the twist: the zero set of a single polynomial is always an algebraic set. However, when we gather many such sets, the combined zero set might not obey the same rules. Like our exercise where the union fails to be an algebraic set. The issue arises since we expect a one-size-fits-all polynomial to match an infinitely varied terrain, which isn’t always possible. This union creates a mathematical landscape that can't be captured by any single polynomial equation, proving that more isn't always merrier in the land of algebraic sets.