CAL1_L5

Theorem 1(Theorem).

Let $S\subset {\mathbb{R}}^n$ such that $\overline{S}={\mathbb{R}}^n$ ($i$.$e$, $S$ is dense in ${\mathbb{R}}^n$). Let $\Omega$ be an open subset of ${\mathbb{R}}^n$. Then, $S\cap \Omega$ is dense in $\Omega$.

Proof
Let $\omega \in \Omega$. If $\omega \in S\cap \Omega$, we are done.
Else, let $N=B_{r_1}(\omega ,{\mathbb{R}}^n)\cap \Omega$ be a neighborhood of $\omega$ in $\Omega$. Since $\Omega$ is open, we know that there exists $r_2$ such that $B_{r_2}(\omega ,{\mathbb{R}}^n)\cap \Omega =B_{r_2}(\omega ,{\mathbb{R}}^n)$ (which is also true for all $r<r_2$). Let $r^{\prime }=\min \{r_1,r_2\}$. Now,

$$\begin{array}{rl}N\cap (\Omega \cap S)& =N\cap S\\ & \subset B_{r^{\prime }}(\omega ,{\mathbb{R}}^n)\cap \Omega \cap S\\ & =B_{r^{\prime }}(\omega ,{\mathbb{R}}^n)\cap S,\end{array}$$

which we know contains a point other than $\omega$ since $S$ is dense in ${\mathbb{R}}^n$.

Theorem 2(Theorem).

Let $f:{\mathbb{R}}^2\to \mathbb{R}$. If ${\lim }_{(x,y)\to (a,b)}f(x,y)=L$ and if the one dimensional limits ${\lim }_{x\to a}f(x,y)$ and ${\lim }_{y\to b}f(x,y)$ both exist, then

$$\underset{x\to a}{\lim }\underset{y\to b}{\lim }f(x,y)=\underset{y\to b}{\lim }\underset{x\to a}{\lim }f(x,y)=L.$$

Proof
Let $g(y)={\lim }_{x\to a}f(x,y)$ and $h(x)={\lim }_{y\to b}f(x,y)$. Let $ϵ>0$. There exists ${\delta }_1>0$ such that 0<\lVert (x, y)-(a, b) \rVert<\delta_{1} \implies$$|f(x, y)-L|<\epsilon.
Now, $0<|x-a|<\frac{{\delta }_1}{\sqrt{2}}$ and $0<|y-b|<\frac{{\delta }_1}{\sqrt{2}}$ implies $0<\sqrt{(x-a{)}^2+(y-b{)}^2}<{\delta }_1$. There also exists ${\delta }^{\prime }>0$ such that 0<|y-b|<\delta'\implies$$|h(x)-f(x, y)|<\epsilon. Let ${\delta }_2=\min \left\{\frac{{\delta }_1}{\sqrt{2}},{\delta }^{\prime }\right\}$.

Finally, $|h(x)-L|\le |h(x)-f(x,y)|+|f(x,y)-L|$. If $0<|x-a|<\frac{{\delta }_1}{\sqrt{2}}$ (and, since the inequality holds for all $y$, we can restrict $0<|y-b|<{\delta }_2$), then $|h(x)-L|\le ϵ+ϵ=2ϵ$. Thus, ${\lim }_{x\to a}h(x)=L$. Ditto for the other iterative limit.