1–8 Theorem. If $A\subset \mathbb R^n$, a function $f:A\to \mathbb R^m$ is continuous if and only if for every open set $U\subset \mathbb R^m$ there is some open set $V\subset \mathbb R^n$ such that $f^{-1}(U) = V\cap A$.
I want to prove this theorem , but i can't understand it even for $n=1$ and $m=1$. I can't see how can a function $\mathbb R\to\mathbb R$ be discontinuous if this theorem is true ,
since we can just take $V = \{0\}$ and then $f^{-1}(U)=\{0\}$ doesn't exist
Take $n=1$ and $m=1$ , and lets define $f:\mathbb R \to\mathbb R$ define by $$f(x)=\begin{cases}x&\text{if } x<2\\ x^2& \text{if }x\geq 2\end{cases}$$
I can't prove that this function isn't continious on $\mathbb R$.
Take the open set $U=]3,3+\frac12[$ , then there is $V=\{0\}$ , such that $f^{-1}(U)=\{0\}$
Best Answer
Note that if $(Y,d|Y\times Y)$ is a subspace of the metric space $(X,d)$, then a set $O'\subseteq Y$ is open (in $Y$) if and only if there exists an open set $O\subseteq X$ such that $O' = O \cap Y$.
The "only if" direction can be proven by considering a set of open balls (in $Y$) $C_\alpha\subseteq O'$, where $\alpha\in C_\alpha$ ($\bigcup_{\alpha\in O'} C_\alpha = O'$); let $B_\alpha$ be the open ball in $X$ with the same radius and center as $C_\alpha$, so that $B_\alpha \cap Y = C_\alpha$. Then, $O' = \bigcup_{\alpha\in O'} (B_\alpha\cap Y) =(\bigcup_{\alpha\in O'} B_\alpha)\cap Y$, so $O = \bigcup_{\alpha\in O'} B_\alpha$ works.
On the other hand, if $O' = O\cap Y$ for some open $O$ (in $X$), then let $O = \bigcup_{\alpha\in O} D_\alpha$ for open balls $D_\alpha$ containing $\alpha$. Then, $D_\alpha \cap Y$ is an open ball (in $Y$), and $\bigcup_{\alpha\in O} (D_\alpha\cap Y) =(\bigcup_{\alpha\in O} D_\alpha)\cap Y = O'$; $O'$ is a union of open balls and thus is open.
Thus, what this theorem reduces to is the fact that $f$ is continuous if and only if the preimage of each open set of the codomain is an open set of $A$, which is a standard definition of continuity.
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