A Higher Perspective on Basic Problems in Calculus

Definiton. Weak topolgy.

Let $X$$X$ be a set and $Y_i$$Y_i$ be a family of topological spaces. Let $(f_i{)}_{i\in I}$$(f_i)_{i\in I}$ be a family of functions from $X$$X$ to $Y_i$$Y_i$, then the weak topolgy on $X$$X$ is defined to be the weakest topology make each $f_i$$f_i$ continuous. It exists since the topology on a Set from a complete lattice. Easy to see that the weak topology is generated by $f_i^{-1}(U)$$f_i^{-1}(U)$ for each $i\in I$$i\in I$ and all the open set $U$$U$ in each $Y_i$$Y_i$.

Universal Property of weak topology.

Let $h:Z\to X$$h:Z\to X$ be a function, then $h$$h$ is continuous iff $f_i\circ h$$f_i\circ h$ is continuous for all $i\in I$$i\in I$.

Proof. $⟹$$\implies$ is obviously. Now let each $f_i\circ h$$f_i\circ h$ be continuous function, then $(f_i\circ h{)}^{-1}(U)=h^{-1}(f_i^{-1}(U))$$(f_i\circ h)^{-1}(U)=h^{-1}(f^{-1}_i(U))$ is open set.

Since $f_i^{-1}(U)$$f_i^{-1}(U)$ generate the weak topology on $X$$X$, $h$$h$ is continuous. $◻$$\square$

Corollary. Let $f:Z\to X,g:Z\to Y$$f:Z\to X,g:Z\to Y$, be two continuous functions, then $(f,g):Z\to X\times Y$$(f,g):Z\to X\times Y$ is continuous.

Proof. Notice that the product topology is the weak topology resepct to ${\pi }_x,{\pi }_y$$\pi_x,\pi_y$, the universal property of weak topology claim that $(f,g)$$(f,g)$ is continuous iff $f,g$$f,g$ is continuous. $◻$$\square$

Proposition.

Let $f,g$$f,g$ be two continuous function from $X\to R$$X\to R$, where $R$$R$ is a topological ring, then $f+g,f\cdot g$$f+g,f\cdot g$ are continuous.

Proof. It follows from $(f,g)$$(f,g)$ is continuous and $+,\cdot$$+,\cdot$ is continuous and the composition of continuous function is continuous directly. $◻$$\square$

$$\begin{array}{}\text{(1)}& X\stackrel{(f,g)}{\to }R\times R\stackrel{+/\cdot }{\to }R\end{array}$$