Prove the set ${\{u \in \mathbb{R^n} | f(u)<c}\}$ is open in $\mathbb{R^n}$.

Question

The following is from the book Advanced Calculus by P.M.Fitzpatrick:

Considering the first case of 4 cases in the Corollary, the two sets ${\{u \in \mathbb{R^n} | f(u)<c}\}$ and ${\{v \in \mathbb{R} | v<c}\}$ are not 'same'. That is the set ${\{v \in \mathbb{R} | v<c}\}$ is an open set but the set ${\{u \in \mathbb{R^n} | f(u)<c}\}$ may not be an open set (it may not be an interval) depending on the defined function $f$.

For example consider ${\{(x,y) \in \mathbb{R^2} | x^2/5+y^2/4<1}\}$ is an open set but $x^2/5+y^2/4<1$ is actually $0\le x^2/5+y^2/4 < 1$. It is a lame example (at least the mentioned Corollary is not useful for this example) but

My question are : 1. Is there exist some fuction $f$ such that $f(u)<c$ to mean $a\le f(u) \le b <c$ becasue of range of the $f$?

And, 2. how ${\{(x,y) \in \mathbb{R^2} | x^2/5+y^2/4<1}\}$ is an open set but $0\le x^2/5+y^2/4 < 1$ is not open in R?

Answer 1

The definition of a continuous function is one for which preimages of open sets are open. The set $\{u\in\mathbb{R}^n|f(u)<c\}$ is the preimage under $f$ of the set $\{v\in\mathbb{R}|v<c\}$, so it must be open by definition of continuity.

If you're working with a different definition of continuity, then you can mimic part of the proof that the definitions are equivalent. To show that preimages of open sets are open, you need to pick a point in the preimage and show that there is some $\delta$-neighborhood around it that is also in the preimage, but that's precisely what the $\epsilon$-$\delta$ definition of continuity states, once we know that the image is open.

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Following the edit of your question, it is certainly the case that the preimage of an open set may also be described as the preimage of some non-open set, but that doesn't harm anything. A simple example, using a function from $\mathbb{R}\to\mathbb{R}$ is this: We can describe $\{x|x^2<1\}$ equally well as $\{x|0\leq x^2<1\}$. Either way, its preimage is the open interval $(-1,1)$.

Additionally: $\{x|-2<\sin x<2\}=\{x|-2\leq\sin x\leq 2\}=\{x|\sin x<2\}$.

Why isn't this a problem? Well, the definition of continuity doesn't say that open sets are preimages only of open sets. It also doesn't say that forward images of open sets must be open. All that is required is IF $C$ is open in the codomain, THEN $f^{-1}(C)$ is open in the domain. That remains true, even if the image of $f$ occupies some part of $C$ that is not, itself, open.

It still won't be the case that an open set (that is not also closed) would be a preimage of a closed set, because an alternative characterization of continuity is that preimages of closed sets are closed.

Answer 2

If $f:X\to Y$ is continuous, then for each open set $B\subset Y$ holds $f^{-1}(B)$ is open in $X$.

Now you have $f:\mathbb R^n\to \mathbb R$ and you consider $B=\{v\in \mathbb R~:~v<c\}$ which is open in $\mathbb R$. Hence $$f^{-1}(B)=\{v\in\mathbb R^n~:~f(v)\in B\}=\{v\in\mathbb R^n~:~f(v)<c\}$$ is open in $\mathbb R^n$.

Answer 3

Here is my understanding. We first recall that $f:\mathbb{R}^n\to \mathbb{R}$ is open if and only if $f^{-1}(U)$ is open in $\mathbb{R}^n$ for any open set in $\mathbb{R}$. If we set $U=\{v\in \mathbb{R}; v<c\}$, then $f^{-1}(U)=\{{\bf u}\in \mathbb{R}^n; f({\bf u})<c\}$. That is why the argument says that the set f^{-1}(U)={{\bf u}\in \mathbb{R}^n; f({\bf u})