explain "open" by examples

Question

I am reading the beginning pages of Lee's Introduction to smooth manifolds. It reads " set Xis open", "inverse map F is open in X", etc.

I am confused about the concept of "open". For example, a function $F: \mathbb{R} \rightarrow \{1\} $, i.e., $f(x)=1$. Is the inverse map $F^{-1}$ open? and why?

Another example is bounded periodic function, say, $F: \mathbb{R} \rightarrow [-1,1] $, i.e., $\cos(x)$. Is the inverse map $\cos^{-1}$ open? and why?

It's great if one explains to me the motivation bedhond the "open" definition. Thanks!

Answer 1

The open sets of a space $X$ are precisely the subsets of $X$ that are elements of the given topology $\mathcal{T}_X$ on $X$.

A topology $\mathcal{T}_X$ on $X$ is a family of subsets of $X$, that is is a subset of the powerset $\mathcal{P}(X)$, that satisfies the axioms of a topology, namely

1. $X,\varnothing \in \mathcal{T}$
2. closed under any union $\bigcup U_i$ of elements $U_i$ of $X$
3. closed under finitely many intersections $\bigcap\limits_i^n U_i$ of elements $U_i$ of $X$

This is first and foremost the axiomatic introduction to what an open set is supposed to be.

Now for $\mathbb{R}^n$ the open sets are precisely what we would expect of an open set to be. For the sake of simplicity consider $\mathbb{R}$ with the standard topology $\mathcal{T}$. The open sets are the open intervalls $(a,b) \subset \mathbb{R}$.

Now one reason why open sets are such a big deal is because the structure preserving maps (i.e. morphisms) in the category of topological spaces are precisely the continous maps.

Let $X, Y$ be topological spaces with their respective topologies $\mathcal{T}_X, \mathcal{T}_Y$. A map $f\colon X\to Y$ is continous if the preimage $f^{-1}(U)$ of any open set $U \in \mathcal{T}_Y$ is open in $\mathcal{T}_X$.

You mentioned something about open inverse functions, i assume that you were most likely be talking about preimages $F^{-1}(U)$ of open sets $U$ under a given function $F$ to be open.

However, there is indeed a notion for open maps. Namely, a function $g\colon X\to Y$ between topological spaces $X,Y$ is open if the image $g(V) \subset Y$ of any open set $V\in X$ is open in $Y$.

Your example of the inverse of the constant map $F\colon \mathbb{R} \to \{1\}$ hardly qualifies since the inverse $F^{-1}\colon \{1\} \to \mathbb{R}$ is not well defined.