Proof regarding continuity and Dirichlet function.

Question

I was given this assignment and I was wondering if my proof makes sense? Also I'd love some feedback and tips on how to improve!
Assume $f(x_{0}) = 0$, we prove $g$ is continuous on $x_{0}$.
Let $ϵ > 0$. We choose δ =1. Let $|x-x_{0}|<δ$, then $left|gleft(xright)-gleft(x_{0}right)right|<ϵ$.
$left|gleft(xright)-gleft(x_{0}right)right|=left|fleft(xright)cdot Dleft(xright)-fleft(x_{0}right)cdot Dleft(x_{0}right)right|=left|fleft(xright)cdot Dleft(xright)-0right|=left|fleft(xright)cdot Dleft(xright)right|<ϵ$
By the density of $Bbb RsetminusBbb Q$ in $Bbb R$, there exists $b∈Bbb R$ such that $x_{0} - 1 < b < x_{0} + 1$ and $b∈Bbb RsetminusBbb Q$. We choose $x = b$, then we have $left|fleft(bright)cdot Dleft(bright)right|=left|fleft(bright)cdot0right|=0<ϵ$
By the density of $Bbb Q$ in $Bbb R$, there exists $c∈Bbb R$ such that $x_{0} - 1 < c < x_{0} + 1$ and $c∈Bbb Q$. We choose x = c, then we have $left|fleft(cright)cdot Dleft(cright)right|=left|fleft(cright)cdot1right|=left|fleft(cright)right|<ϵ$
Note that f is continuous at $x_{0}$, therefore $left|fleft(cright)-fleft(x_{0}right)right|<ϵ$. $left|fleft(cright)right|=left|fleft(cright)-fleft(x_{0}right)+fleft(x_{0}right)right|leleft|fleft(cright)-fleft(x_{0}right)right|+left|fleft(aright)right|=left|fleft(cright)-fleft(x_{0}right)right|+left|0right|=left|fleft(cright)-fleft(x_{0}right)right|<ϵ$
Therefore $g$ is continuous at $x_{0}$.
Assume $g$ is continuous at $x_{0}$. We prove $f(x_{0})=0$. Note that f is continuous on $a$. Assume toward contradiction that $f(x_{0})≠0$. Then we have $gleft(x_{0}right)=fleft(x_{0}right)cdot Dleft(x_{0}right) ➜ frac{gleft(x_{0}right)}{fleft(x_{0}right)}=Dleft(x_{0}right)$
By the algebra of continuous functions $frac{gleft(x_{0}right)}{fleft(x_{0}right)}$ is continuous, but we proved in class that $D(x)$ is not continuous on any $x∈Bbb R$. That means we have a contradiction, since we have a continuous function on one side of the equation, and a discontinuous function on the other which can't be.
Therefore $f(x_{0}) = 0$.

azif00 · Answer

Assume that $f(x_0) = 0$, and let $varepsilon>0$. Then we need to find a $delta>0$ such that for each $x in (x_0-delta,x_0+delta)$ the following holds: $|f(x)D(x)| 0$ such that $|f(x)| = |f(x)-f(x_0)|0$ and the continuity of $f$ at $x_0$ gives us a $delta>0$ such that $|f(x)-f(x_0)|<|f(x_0)|/2$ for all $x in (x_0-delta,x_0+delta)$; which implies that $|f(x)| > |f(x_0)|/2$ for all $x in (x_0-delta,x_0+delta)$. Thus, $f(x) neq 0$ for all $x in (x_0-delta,x_0+delta)$, and then we can define $h : (x_0-delta,x_0+delta) to mathbb R$ by $$forall x in (x_0-delta,x_0+delta) : quad h(x) = frac{g(x)}{f(x)}.$$ Now, by the continuity of $f$ and $g$ at $x_0$, $h$ is also continuous at $x_0$, but $h$ is the same as the restriction of $D$ to $(x_0-delta,x_0+delta)$, a contradiction since $D$ is discontinuous in all $mathbb R$.

Proof regarding continuity and Dirichlet function.

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