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

Mathematics Asked on December 3, 2021

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.

One Answer

Yes, by what you posted in your definitions they're correct. Now if you know how to prove that for the product then for the quotient is the same thing. You only take $c=frac{a}{b}=frac{1}{b}cdot a$. That is, you see it like a product.

Answered by Iesus Dave Sanz on December 3, 2021

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