Does removing an open set from topology in $\mathbb{R}$ and associated open sets leads to $\mathbb{R}$ having trivial topology?

Question

Suppose we have $\mathbb{R}$ with the standard topology, and we remove a ball $(a- \epsilon, a+\epsilon)$ from the topology, then what topology do we get, provided we remove only the minimum number of sets to keep it a topology?

This is a thought experiment I came up with, so the wording may be tricky. I'll try to explain what I mean. Suppose we had the set $ \{ 0,1,2 \} $ with the following topology $$\tau = \{ \varnothing, \{ 0 \} , \{1 \} , \{1, 0 \} ,\{ 0 ,1,2 \} \}$$

Now, if I were to remove the element $\{1,0 \}$, and were to end up with a topology after removal, then I would have to neccesarily remove either $\{1 \}$ or $\{0 \}$. It could also be that we remove both, but then we would be violating the "minimum removal condition".

I believe that if we do a process for $\mathbb{R}$ , then we would neccesarily end up with the trivial topology. My intuition is based on the fact that the epsilon interval about $a$ I wrote, can be written by intersection of many other pairs of intervals intervals, now these intervals can be written as intersection of infinitely many other pairs of intervals. Going on, we see we must remove all the open sets completely.

How do I show this process would terminate/does not?

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Difficulties in this question:

Well, I guess, to properly answer this question, one would have to explain have to formalize the idea of the intuitive removal procedure. Which is not so clear to do, hence I tagged as application.

After some discussion offsite, I find that the result of a removal process can just be rephrased into finding "maximal sub topology which doesn't contain an interval".

If the problem can be solved in other formalizations, I welcome those as solutions too.

Answer 1

There are non-trivial topologies that are coarser than the standard topology and do not contain $(a-\epsilon, a + \epsilon)$. This implies that any maximal such topology cannot be trivial. In fact there isn't a unique maximal topology under these constraints.

Given $[b, c] \subset \mathbb{R}$, consider the collection $$\tau_{[b,c]} = \{U \subset \mathbb{R} \mid U \text{ open, either } [b, c] \subset U \text{ or } [b,c] \cap U = \emptyset\}.$$ It is easy to directly check that this is a topology. Alternatively, $\tau_{[b,c]}$ is the pullback of the standard topology on $\mathbb{R}$ under the continuous map $f_{[b,c]} : \mathbb{R} \to \mathbb{R}$, $$f_{[b,c]}(x) = \begin{cases} x - b & \text{if } x < b \\ 0 & \text{if } x \in [b,c] \\x - c & \text{if } x > c\,. \end{cases}$$ In other words, $$\tau_{[b,c]} = \{f_{[b,c]}^{-1}(U) \mid U \subset \mathbb{R} \text{ open}\} \,,$$ which also shows that $\tau_{[b,c]}$ is a topology (it is the coarsest such that $f_{[b, c]}$ is continuous, in particular coarser than the standard topology).

Now, given $a$ and $\epsilon > 0$, consider $\tau_- = \tau_{[a-\epsilon, a-\epsilon/2]}$ and $\tau_+ = \tau_{[a+\epsilon/2, a + \epsilon]}$. Then directly from definition, neither $\tau_-$ nor $\tau_+$ contains $(a-\epsilon, a + \epsilon)$. But if $\tau \supset \tau_- \cup \tau_+$ is a topology then $$(a- \epsilon, a + \epsilon) = (a - \epsilon, a+\epsilon/2) \cup (a-\epsilon/2, a + \epsilon) \in \tau$$ since $(a - \epsilon, a+\epsilon/2) \in \tau_+$ and $(a-\epsilon/2, a + \epsilon) \in \tau_-$.

In particular the set $$\{\tau \text{ topology on \mathbb{R}}\mid \tau \subset \tau_{\text{std}}, (a-\epsilon, a +\epsilon) \notin \tau\}$$ cannot have a single largest (under $\subset$) element $\tau_*$ that we could define as the topology obtained by removing $(a- \epsilon, a+ \epsilon)$. In other words, there is no unique choice of "associated open sets" as in the question title.