Let $\triangle=\{(t_0,t_1,t_2)\in \mathbb{R^3}:t_0+t_1+t_2=1$ and $t_i\geq0$ for $i=0,1,2\}$. Prove that the function $f:[0,1]\times [0,1]\longrightarrow \triangle$ defined by $f(x_1,x_2) = \left\{ \begin{array}{ll} (x_1,x_2-x_1,1-x_2) & \mbox{if } x_1 \leq x_2 \\ (x_2,x_1-x_2,1-x_1) & \mbox{if } x_2 \leq x_1 \end{array} \right.$ is continuous.
$My Approach:$ I have use the sequential criterion to prove this. Choose $(c_1,c_2)\in [0,1]\times [0,1] $ arbitrary. Take any sequence $(x_n,y_n)$ converging to$(c_1,c_2)$. This implies $x_n\longrightarrow c_1$ and $y_n\longrightarrow c_2$ as $n\longrightarrow \infty$.This will imply that $f(x_n,y_n)\longrightarrow f(c_1,c_2)$. Hence $f$ is continuous.
$Question:$ Is my approach is correct? Also, we need to prove $f(A\times B)$ is closed if $A$ and $B$ are closed subset of $[0,1]$. Please Help.
Let $U=\{(x,y)\in[0,1]^2|x<y\}$, $V=\{(x,y)\in[0,1]^2|x>y\}$ and $D=\{(x,x)| x\in[0,1]\}$
The function $f$ is continuous on $U$ and $V$ because the coordinate functions of the restrictrions $f_{|U}$ and $f_{|V}$ are polynomial.
Let $x\in[0,1]$. Show the continuity of $f$ at $(x,x)$
Let $\varepsilon>0$. Let's define $\delta=\frac{\varepsilon}{3}$.
Let $(x_1,x_2)\in [0,1]^2$ such that $(x_1,x_2) \in\mathcal{B}((x,x),\delta)$. We have
$\sqrt{(x_1-x)^2+(x_2-x)^2}<\delta$, thus
$|x_1-x|\leq \sqrt{(x_1-x)^2+(x_2-x)^2}<\delta$ and $|x_2-x|<\delta$
Moreover $|x_1-x_2| \leq |x_1-x+x-x_2|\leq |x_1-x|+|x-x_2|<2\delta$
If $(x_1,x_2)\in U$ then
$||f(x_1,x_2)-f(x,x)||=||(x_1-x,x_2-x_1,-x_2+x)||=\sqrt{(x_1-x)^2+(x_2-x_1)^2+(x-x_2)^2}$
$\leq \sqrt{\delta^2+4\delta^2+\delta^2}\leq \sqrt{6}\delta<\varepsilon$
idem if $(x_1,x_2)\in V$ (by switching $x_1$ and $x_2$)
To prove that $f(A\times B)$ is closed, we can use the theorem of the image of a compact by a continuous function.