Question

Let \(A\) be any nonempty set, and let \(\mathcal{F}_{A}\) be the set of all functions from \(A\) to \(\mathbb{R}\). (a) Why is \(F+G \in \mathcal{F}_{A}\) for all \(F, G \in \mathcal{F}_{A}\). (b) Prove \((F+G)+H=F+(G+H)\) for all \(F, G, H \in \mathcal{F}_{A}\) (c) Let Zero e \(\mathcal{F}_{A}\) be defined by \(\operatorname{Zero}(a)=0\) for all \(a \in A\). Prove that \(Z e r o+F=F\) for all \(F \in \mathcal{F}_{A}\) (d) For \(F \in \mathcal{F}_{A}\), define \(\bar{F}\) by \(\vec{F}(a)=-F(a)\) for each \(a \in A\). Prove that \(F+\bar{F}=\) Zero \(=\bar{F}+F\) for all \(F \in \mathcal{F}_{\mathcal{A}}\)

Short answer

(a) Functions are closed under addition. (b) Addition is associative. (c) Zero function is an identity element. (d) \(\bar{F}\) is the additive inverse of \(F\).

Step-by-step solution

Understanding the Set of Functions

Given a nonempty set \(A\), \(\mathcal{F}_A\) represents the set of all functions from \(A\) to \(\mathbb{R}\). This means any function \(F\) in \(\mathcal{F}_A\) assigns a real number to each element of \(A\).

(a) Closure Under Addition

To show that \(F + G \in \mathcal{F}_A\) for any functions \(F, G \in \mathcal{F}_A\), note that for each \(a \in A\), we have \((F + G)(a) = F(a) + G(a)\). Since both \(F(a)\) and \(G(a)\) are real numbers (because \(F, G\) are functions from \(A\) to \(\mathbb{R}\)), \(F(a) + G(a)\) is also a real number. Thus, \(F + G\) is a function from \(A\) to \(\mathbb{R}\), ensuring closure under addition.

(b) Associativity of Addition

To prove \((F+G)+H=F+(G+H)\), consider any \(a \in A\). We compute:\[(F+G)(a) + H(a) = (F(a) + G(a)) + H(a),\]\[F(a) + (G(a) + H(a)) = F(a) + (G+H)(a).\]By the associativity of real number addition, we have \((F(a) + G(a)) + H(a) = F(a) + (G(a) + H(a))\). This holds for every \(a \in A\), proving that function addition is associative.

(c) Additive Identity

To show that \(\text{Zero} + F = F\), consider any \(a \in A\). We have:\[(\text{Zero} + F)(a) = \text{Zero}(a) + F(a) = 0 + F(a) = F(a).\]Since this equality holds for all \(a \in A\), \(\text{Zero} + F = F\) for all \(F \in \mathcal{F}_A\). This means \(\text{Zero}\) is the additive identity in \(\mathcal{F}_A\).

(d) Additive Inverses

To prove \(F + \bar{F} = \text{Zero}\), take any \(a \in A\) and calculate:\[(F + \bar{F})(a) = F(a) + \bar{F}(a) = F(a) + (-F(a)) = 0 = \text{Zero}(a).\]Similarly, \(\bar{F} + F = \text{Zero}\) because:\[(\bar{F} + F)(a) = \bar{F}(a) + F(a) = (-F(a)) + F(a) = 0 = \text{Zero}(a).\]Thus, \(F + \bar{F} = \text{Zero} = \bar{F} + F\), indicating that \(\bar{F}\) is the additive inverse of \(F\) in \(\mathcal{F}_A\).

Key concepts

Closure Under Addition

When we talk about closure under addition, we're dealing with a very important mathematical property of sets with an operation defined on them. Imagine a big box labeled

• Functions: From a set \(A\) to real numbers

Now, if you pick out any two functions from this box and add them together, the result should still fit inside the box.
This is what closure under addition guarantees: that the sum of any two functions, say \(F\) and \(G\), remains a function from \(A\) to \(\mathbb{R}\). If you have a function \(F\) that maps each element \(a\) from set \(A\) to a real number \(F(a)\), and similarly, function \(G\) that maps \(a\) to \(G(a)\), then their sum \(F+G\) defines a new function. This new function maps \(a\) to the sum \(F(a) + G(a)\), which is still a real number. So, closure ensures everything stays nice and tidy inside the big box of functions from \(A\) to \(\mathbb{R}\).

Associativity

Associativity of addition is like having a friendly agreement about the order in which you add numbers. It doesn't matter how you group them, as the result will be the same.
When applied to functions, this means if you have three functions \(F\), \(G\), and \(H\), the way you add them together doesn't change the outcome. Say we want to add them and we start by adding

• \(F + G\)

first and then add \(H\), or if we first add

• \(G + H\).

Either way, you get the same final function. Mathematically, this is written as:

• \((F+G)+H = F+(G+H)\).

For each element \(a\) in \(A\), the associative property ensures that:

• \((F(a) + G(a)) + H(a) = F(a) + (G(a) + H(a))\)

thanks to the associativity of real numbers.
It's like picking up three smoothie drinks and mixing them; no matter the glass sequence, the flavor remains constant.

Additive Identity

Think of an additive identity as the "zero" of combining things - it doesn't change anything you add it to. For functions mapping from \(A\) to \(\mathbb{R}\), there's a special function called "Zero". This function assigns each element in \(A\) the number 0.
The reason \(Zero\) holds the title of additive identity is because it doesn't alter any function you add it to. For a function \(F\), when you add \(Zero\) to \(F\), for any element \(a\) in \(A\), it holds that:

• \((\text{Zero} + F)(a) = \text{Zero}(a) + F(a) = 0 + F(a) = F(a)\).

It's like having a clear sheet among colored sheets of paper; when put together, it doesn't change the color, just as 0 added to any number keeps the original number intact.
So, \(Zero\) acts as the neutral party within the set of functions \(\mathcal{F}_{A}\).

Additive Inverses

An additive inverse is like the perfect opposite of an element, in this case, a function, that when added together results in a neutral outcome, or zero. For any function \(F\), there's another function \(\bar{F}\) that acts as its twin with opposite traits.
The way \(\bar{F}\) is constructed is by flipping every output of \(F\) to its negative counterpart:
for every \(a\) in \(A\), \(\bar{F}(a)=-F(a)\).
When you add \(F\) and \(\bar{F}\), for each \(a\), you compute:

• \((F + \bar{F})(a) = F(a) + \bar{F}(a) = F(a) + (-F(a)) = 0\).

This means that the pair \(F\) and \(\bar{F}\) come together to give the Zero function, which maps every element of \(A\) to 0.
It's like wearing and removing a hat – together, they bring back your original look, much as \(F + \bar{F}\) restores the additive identity.