I am a bit confused on how to prove that $\mathbb{R}^2 \rightarrow \mathbb{R}$ sending $(x,y)$ to $x+y$ and $xy$ are continuous. I am currently reading Hatcher's point set topology notes. My proof so far is,
Let $(\mathbb{R},d)$ be a metric space where $d$ is euclidean metric. This metric space induces a topological space labeled $(\mathbb{R},\mathcal{O})$. We can define a topology on $\mathbb{R}^2$ by defining the basis $\mathcal{B} = \{U \times V: U,V \in \mathcal{O} \}$. This is a basis since it satisfies (1) and (2) in Hatcher(Proposition 2). Since $\mathbb{R} \in \mathcal{O}$, then $\mathbb{R}^2 \in \mathcal{B}$ which satisfies (1). To show that it satisfies (2), suppose $U_1 \times V_1 \in \mathcal{B}$ and $U_2 \times V_2 \in \mathcal{B}$. Let, $x \in (U_1 \times V_1) \cap (U_2 \times V_2)$. Since $(U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2)$ and $\mathcal{O}$ is closed under finite intersections, then there exists $B_3 = (U_1 \cap U_2) \times (V_1 \cap V_2) \in \mathcal{B}$ such that $x \in B_3 \subset (U_1 \cap U_2) \times (V_1 \cap V_2)$. Thus, $\mathcal{B}$ is a basis for a topology on $\mathbb{R}^2$ denoted as $\mathcal{O}_2$.
I want to show that both functions($f(x,y) = x+y$ and $g(x,y) = xy$) are continuous(note, $(\mathbb{R},\mathcal{O})$ and $(\mathbb{R}^2,\mathcal{O}_2)$. I suppose that $O \in \mathcal{O}$. I am trying to show that $f^{-1}(O) \in O_2$. I am struggling with what $f^{-1}(O)$ set looks like.
 A: In Hatcher's notes, this is exercise 13 p. 15, which requires "using
only definitions and results from this class, not results from calculus for example". On p. 14, Hatcher advised you to "check that [your map $f:(x,y)\mapsto x+y$] is continuous by seeing directly that the inverse image of an open interval is open". On p. 12 he said that the product topology on $\mathbb R^2$ coincides with the "usual topology", which he defined on p. 3.
Let $U=f^{-1}((a,b))$ and $(x_0,y_0)\in U$, i.e. $a<x_0+y_0<b$. Let $\epsilon=\frac12\min(b-x_0-y_0,x_0+y_0-a)$. Then, the open square $(x_0-\epsilon,x_0+\epsilon)\times(y_0-\epsilon,y_0+\epsilon)$ is contained in $U$. This proves that $U$ is open, q.e.d.
A: The continuity of $\phi : X \to Y$ can be characterized by various equivalent properties. The standard definition is probably that $\phi^{-1}(V)$ is open in $X$ whenever $V \subset Y$ is open. But sometimes it is easier to work with this property:

$\phi$ is continuous iff it is continuous at each $x \in X$.  Continuity at $x \in X$ means that for each open neigbborhood $V$ of $\phi(x)$ in $Y$ there exists an open neigborhood $U$ of $x$ in $X$ such that $\phi(U) \subset V$.

Let $(x,y) \in \mathbb R^2$ and $V$ be an open neigborhood of $s = f(x,y) = x+y$ in $\mathbb R$. Choose $r >0$ such that $(s-r,s+r) \subset V$. Then $U = (x-r/2,x+r/2) \times (y-r/2,y+r/2)$ is an open neighborhood of $(x,y)$ in $\mathbb  R^2$ such that $f(U) \subset (s-r,s+r) \subset V$.
Let $(x,y) \in \mathbb R^2$ and $V$ be an open neigborhood of $p = g(x,y) = x \cdot y$ in $\mathbb R$. Choose $r >0$ such that $(p-r,p+r) \subset V$. Choose $u> 0$ such that $u^2 < r/2$ and $u(\lvert x \rvert + \lvert y \rvert) < r/2$. Then $U = (x-u,x+u) \times (y-u,y+u)$ is an open neighborhood of $(x,y)$ in $\mathbb  R^2$ such that $g(U) \subset (p-r,p+r) \subset V$.
A: Another method of proving maps are continuous is to show they preserve limits of sequences. That is if:
$$f(\lim\limits_{n\in\mathbb{N}}s_n) = \lim\limits_{n\in\mathbb{N}}f(s_n),$$
for an arbitrary sequence $\{s_n\}_{n\in\mathbb{N}}\in dom(f)$.

If we argue that any sequence of points in $\mathbb{R}^2$ is given by a collection of coordinates indexed as follows:
$$\{(x_n,y_n)\}_{n\in\mathbb{N}}$$
Then with the addition and multiplication maps, we have respectively:
$$\color{blue}{+\big(\lim\limits_{n\to\infty}(x_n,y_n)\big)} = +\big((\lim\limits_{n\in\mathbb{N}}x_n,\lim\limits_{n\in\mathbb{N}}y_n)\big):= \lim\limits_{n\in\mathbb{N}} x_n + \lim\limits_{n\in\mathbb{N}}y_n = \lim\limits_{n\in\mathbb{N}}(x_n+y_n) =: \color{blue}{\lim\limits_{n\in\mathbb{N}}+\big((x_n,y_n)\big)}$$
and
$$\color{blue}{*\big(\lim\limits_{n\to\infty}(x_n,y_n)\big)} = *\big((\lim\limits_{n\in\mathbb{N}}x_n,\lim\limits_{n\in\mathbb{N}}y_n)\big):= \lim\limits_{n\in\mathbb{N}} x_n * \lim\limits_{n\in\mathbb{N}}y_n = \lim\limits_{n\in\mathbb{N}}(x_n*y_n) =: \color{blue}{\lim\limits_{n\in\mathbb{N}}*\big((x_n,y_n)\big)}.$$
The limit applying component wise in the first equality is more or less intuitive. The compatibility with limits in the 3rd equalities is assumed known.
