The weak topology for $\mathbb{R}$ generated by some functions

general-topologyreal-analysisweak-topology

I am reading these two question one and two, and thinking about that question:

We assume that $(\varphi_i)$ is a family of real valued functions, i.e. $\varphi:\mathbb{R}\to \mathbb{R}$. So for every item below what is the weak topology generated by:

  1. $(\varphi_i)$ is all constant fuctions
  2. $(\varphi_i)$ is only one function
    $$\varphi(x) = \begin{cases}
    0, & \text{if } x \leq 0,\\
    1, & \text{if } x > 0.
    \end{cases}
    $$
  3. $(\varphi_i)$ are all limited functions such that is continuous with respect to $\mathbb{R}$ usual topology.
  4. $(\varphi_i)$ are all functions that are continuous with respect to the usual topology of $\mathbb{R}$.

Thinking about this I wrote

  1. The topology $\sigma$ on $\mathbb{R}$ is by definition the coarsest topology on $\mathbb{R}$ such that all constant function $\varphi_i : \mathbb{R} \to \mathbb{R}$ is continuous when the domain $\mathbb{R}$ is equiped with $\sigma$ and the codomain is the usual topology.

    We know that every constant function is continuous, since $\mathbb{R}$ is open and, given an open $A$ which contains $c$ where $\varphi(x)=c$ for every $x \in \mathbb{R}$, we have that $f^{-1}(A) = \mathbb{R}$. So the thickest topology is the indiscreet topology, i.e. $\sigma = \{\varnothing, X\}$.

  2. This function will generate a weak topology?

  3. We have that every interval of type $(a,b)$ belongs to the weak topology $\sigma$. This is because we have a limited function of the type
    $$
    f(x) = \begin{cases}
    -1, & \text{if } x < a,\\
    \frac{2x-(a+b)}{b-a} & \text{if } a \leq x \leq b,\\
    1, & \text{if } x > b.
    \end{cases}
    $$
    and $f^{-1}((-1,1)) = (a,b)$. So, since the interval $(-1,1)$ is open in the usual topology, the interval $(a,b)$ must be open in $\sigma$, so $(a,b) ) \in \sigma$.

Conversely, every open set in $\sigma$ is of the form $(a,b)$ and belongs to the usual topology, so $\sigma$ is equal to the usual topology.

  1. The weak topology will be the usual topology on $\mathbb{R}$?

I think my doubts are numbers 2 and 4 of that list. How those weak topologies look?

Thanks for any help 🙂

Best Answer

Answers to 2) and 4):

  1. The topology is $\{\emptyset, \mathbb R, (-\infty,0], (0,\infty)\}$. [Note that $\phi^{-1} (-1,1/2)=(-\infty,0]$ and $\phi^{-1} (1/2,2)=(0,\infty$].

  2. This is just the usual topology. Let $\tau'$ be the toplogy generated by the given functions. Since the identity function is continuous w.r.t. the usual topology $\tau$ it follows that $\tau \subseteq \tau'$. Since $\tau'$ is defined as the smallest topology which makes continuous functions w.r.t. $\tau$ continuous, it follows that $\tau'\subseteq \tau$.

Related Question