Prove that the parametric points set $\{(f(t), g(t)) \mid t \in \mathbb{R} \}$ is a straight line if $f, g$ are affine functions Let be $L=\{(f(t), g(t)) \mid t \in \mathbb{R} \} \subseteq \mathbb{R}^2$, where $f, g$ are applications $f, g \colon \mathbb{R} \to \mathbb{R}$. Prove that:

*

*if $f, g$ are continuous functions, then $L$ is a line

*if $f, g$ are affine functions ($x \mapsto ax+b$ for some $a,b$), then $L$ is a straight line

Intuitively, I see that, but I don't know how to prove, how to formally define a line and straight line (I only know Jordan theorem: "a closed curve splits the plane in two arc-connex areas"; but I don't know formal definition about straight line)
 A: Regarding the second statement: In order to show that $L$ is a straight line in $\mathbb{R}^2$, it suffices to show that the graph of the corresponding affine function of $L$ has constant slope, since said affine function will take the form $h(x) = \alpha x+\beta$ for $\alpha,\beta \in \mathbb{R}$. Let $f(t) = at+b$ and $g(t) = ct+d$ with $a,b,c,d \in \mathbb{R}$. For arbitrary fixed $t_1,t_2 \in \mathbb{R}$ we check the slope by the standard formula "rise over run", given by
\begin{align}
\alpha = \frac{y_2-y_1}{x_2-x_1}.
\end{align}
In our case we have $y_i = g(t_i)$ and $x_i = f(t_i)$ for $i=1,2$. We thus have
\begin{align}
\alpha = \frac{g(t_2)-g(t_1)}{f(t_2)-f(t_1)} = \frac{ct_2+d-ct_1-d}{at_2+b-at_1-b} = \frac{ct_2-ct_1}{at_2-at_1} = \frac{c(t_2-t_1)}{a(t_2-t_1)} = \frac{c}{a},
\end{align}
so the slope is always constant, since $a$ and $c$ were fixed. Hence $L$ is a straight line.
I can't help you with the first statement, the only definition of "line" I know is that of a straight line.
Hope this helps!
Edit: I'm not sure how much you know about topology, so this might not be very helpful, but if you define "line" as simply a continuous path in $\mathbb{R}^2$, I have the following answer:
Let $\mathcal{O}$ denote the standard topology on $\mathbb{R}$. The set $B := \{ U \times V \mid U,V \in \mathcal{O} \}$ is a base for the standard topology on $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$. We then endow $L \subseteq \mathbb{R}^2$ with the subspace topology, so $B_L := \{ L \cap (U \times V) \mid U,V \in \mathcal{O} \}$ is a base for this topology. Define
$$
F : \mathbb{R} \to \mathbb{R}^2, \quad t \mapsto (f(t),g(t)).
$$
Now our goal is to show that $F$ is continuous, because then $L$ is a continuous path in $\mathbb{R}^2$. In order to check that $F$ is indeed continuous, we only need to show that the preimage of a single element of $B_L$ is open in $\mathbb{R}$ (see for example this post).
Let $\text{pr}_{x,y} : \mathbb{R}^2 \to \mathbb{R}$ be the canonical projections, i.e. $\text{pr}_x(x_0,y_0)=x_0$ and similarly with $\text{pr}_y$. Now, take an arbitrary element from $B_L$, say $L \cap (U \times V)$ with $U,V \in \mathcal{O}$. We compute
\begin{align}
F^{-1}(L \cap (U \times V)) &= F^{-1}((\text{pr}_x(L) \cap U) \times (\text{pr}_y(L) \cap V)) \\
&= f^{-1}(\text{pr}_x(L) \cap U) \cap g^{-1}(\text{pr}_y(L) \cap V).
\end{align}
Now, we see that $\text{pr}_x(L) \cap U \in \mathcal{O}$ and similarly $\text{pr}_y(L) \cap V \in \mathcal{O}$, since the projection is an open map (since $f,g$ are continuous). But since $f$ and $g$ are continuous, we also get that their preimages under $f$ and $g$ respectively are open in $\mathbb{R}$. So we have shown that $F^{-1}(L \cap (U \times V)) \in \mathcal{O}$.
