Consider the function $f:(\mathbb{R},T)\rightarrow(\mathbb{R},T)$ such that $f(x)=x^2$ for all $x$ in $\mathbb{R}$ and $T=\{\emptyset,\mathbb{R}\}\cup\{(-\infty,x):x\in\mathbb{R}\}$. Prove that $f$ if not continuous at $a>0$.
OK first of all I do not get where the '$a$' comes from. Anyways, can anyone help me.
Let $a,b\in \Bbb R^+.$ Then $a\in (-\infty, a+b)\in T.$ If $f$ were continuous at $a$ then for some $t\in T$ we would have $a\in t$ and $\{f(x):x\in t\} \subset (-\infty,a+b).$
But if $a\in t\in T$ then for some $c\in \Bbb R^+$ we have $t\supset (\infty,a+c)\supset (-\infty,0)$ so $$\{f(x):x\in t\}\supset \{f(x):x<0\}=\{x^2:x<0\}=\Bbb R^+\not \subset (-\infty,a+b).$$