Proving that $f + g, f cdot g$, and $f/g$ are continuous using neighborhoods

This is Exercise 1 from Section 1.2, page 9, of Topology and Groupoids by Brown. I paraphrase portions of the book to hopefully improve clarity.

I found questions such as Continuity of the sum of continuous functions and Topology: Continuous Functions but they look a bit too advanced and don’t quite address my problems.

Exercise:

Let $f colon A to mathbb{R}$ and $g colon B to mathbb{R}$ be functions where $A$ and $B$ are subsets of $mathbb{R}$. Prove that if $f$ and $g$ are continuous at $a$, then $f + g$ (on domain $A cap B$), $f cdot g$ (on domain $A cap B$), and $frac{f}{g}$ (on domain $(A cap B) setminus {x in B colon g(x) = 0 }$) are all continuous at $a$ (assuming $a$ lies within the respective domains).

I am supposed to use the following results in my proofs (from Exercises 6 and 7 on page 4 of the same book):

Let $C$ be a neighborhood of $c in mathbb{R}$, and let $a + b = c$. Then there are neighborhoods $A$ of $a$ and $B$ of $b$ such that $x in A$ and $y in B$ imply $x + y in C$.

If $c = ab$ then there are neighborhoods $A^{prime}$ of $a$ and $B^{prime}$ of $b$ such that $x in A^{prime}$ and $y in B^{prime}$ imply $xy in C$.

Other information:

The definition of continuity I have decided to use is Definition $C^{prime}$ from page 4 of the same book. It says that the function $f colon A to mathbb{R}$ is continuous at $a in A$ if for each neighborhood $N$ of $f(a)$, there is a neighborhood $M$ of $a$ such that for all $x in mathbb{R}, x in M cap A Rightarrow f(x) in N.$

Attempt for $f + g$:

Consider some neighborhood $N$ of $(f + g)(a)$. We want to show that there is a neighborhood $M$ of $a$ such that for all $x in mathbb{R}, x in M cap A cap B Rightarrow (f + g)(x) in N$.

Using the additive result mentioned, we know that given $N$ there are neighborhoods $N^{prime}$ of $f(a)$ and $N^{prime prime}$ of $g(a)$ such that $y in N^{prime}$ and $z in N^{prime prime}$ imply $y + z in N$.

By continuity of $f$ at $a$, we know that given $N^{prime}$, there is a neighborhood $M^{prime}$ of $a$ such that for all $alpha in mathbb{R}, alpha in M^{prime} cap A cap B Rightarrow f(alpha) in N^{prime}$.

By continuity of $g$ at $a$, we know that given $N^{prime prime}$, there is a neighborhood $M ^{prime prime}$ of $a$ such that for all $beta in mathbb{R}, beta in M^{prime prime} cap A cap B Rightarrow g(beta) in N^{prime prime}$.

If we let $M = M^{prime} cap M^{prime prime}$, then for all $x in mathbb{R}, x in M cap A cap B$ implies $f(x) in N^{prime}$ and $g(x) in N^{prime}$, which implies that $f(x) + g(x) = (f + g)(x) in N$ as desired.

Attempt for $f cdot g$:

This proof proceeds the same way as the prior one, except that we replace all instances of $+$ with $cdot$ and rely on the multiplicative result mentioned.

Questions:

Are my two proofs above correct?

How should I prove continuity of $frac{f}{g}$? I do not understand how to use the multiplicative result mentioned to help me here. Do I need to prove a similar result for $c = frac{a}{b}$?

Thanks.

Edit:

After some thought, here is my attempt for $frac{f}{g}$. I use Section 1.1 Exercise 8 from the text, which gives the result that if $C$ is a neighborhood of $c$, where $c neq 0$, then there is a neighborhood $C^{prime}$ of $frac{1}{c}$ such that if $x$ is in $C^{prime}$, then $frac{1}{x}$ is in $C$. My attempt:

Looking at the multiplicative result, there is no reason we cannot consider the case when $c = a cdot frac{1}{b}$ since after all it’s still just the product of two numbers.

Consider some neighborhood $N$ of $big( frac{f}{g} big) (a)$. By the multiplicative result there must be neighborhoods $N^{prime}$ of $f(a)$ and $N^{prime prime}$ of $frac{1}{g(a)}$ such that $y in N^{prime}$ and $frac{1}{z} in N^{prime prime}$ imply $frac{y}{z} in N$.

By continuity of $f$ at $a$, given $N^{prime}$ there is a neighborhood $M^{prime}$ of $a$ such that for all $alpha in mathbb{R}, alpha in M^{prime} cap (A cap B) setminus {x in B colon g(x) = 0 } Rightarrow f(alpha) in N^{prime}$.

For $g$, we need to call on Exercise 8 mentioned above. We have a neighborhood $N^{prime prime}$ of $frac{1}{g(a)}$. Exercise 8 tells us that there is a neighborhood $N^{prime prime prime}$ of $g(a)$ such that if $z in N^{prime prime prime}, frac{1}{z} in N^{prime prime}$. By continuity of $g$ at $a$, given $N^{prime prime prime}$ there is a neighborhood $M^{prime prime prime}$ of $a$ such that for all $beta in mathbb{R}, beta in M^{prime prime prime} cap (A cap B) setminus {x in B colon g(x) = 0 } Rightarrow g(beta) in N^{prime prime prime}$. Since $g(beta) in N^{prime prime prime}$, we must have $frac{1}{g(beta)} in N^{prime prime}$.

Let $M = M^{prime} cap M^{prime prime prime}$. Then for all $x in mathbb{R}, x in M cap (A cap B) setminus {x in B colon g(x) = 0 }$ implies $f(x) in N^{prime}$ and $frac{1}{g(x)} in N^{prime prime}$, which implies that $frac{f(x)}{g(x)} = big( frac{f}{g} big) (x) in N$ as desired.