# Are these two sets clopen and closed?

Let $$F=\{(-\infty, x): x\in \mathbb{R} \} \ \cup \{ \emptyset, \mathbb{R}\}$$. Show that $$F$$ is a topology and describe the closure of $$(0,1)$$ and the interior of $$[0,1]$$ in this topology.

I managed to show that $$F$$ is indeed a topology but I'm a little unsure of the second part. We define a closed set as a set whose complement is open in this case $$(0,1)$$ is closed if $$\{\emptyset,\mathbb{R}\}$$ is open, I know that in a topological space both the empty set and the whole space is considered clopen so I think the intervall must also be clopen and from that I assume that $$[0,1]$$ is just closed . I think my thought process is solid but I'm not entirely sure so I would appreciate some clarification.

• $\{\varnothing,\mathbb{R}\}$ is not a subset of the reals. It makes no sense to say,conditionally or not, that it is "open" in this topology. Oct 29, 2022 at 0:01

Closure of $$(0,1)$$ is $$[0,\infty)$$. Just check that every neighborhood of $$x$$ contains points of $$(0,1)$$ for $$x \geq 0$$ and that $$x<0$$ implies $$(-\infty, y)$$ is a neighborhood of $$x$$ containing no point of $$(0,1)$$ if $$x.
No bounded set has any interior points so interior of $$[0,1]$$ is empty.
$$F$$ is a topology defined on $$\mathbb{R}$$. The elements in $$F$$, which are called the open sets of $$F$$, are subsets of $$\mathbb{R}$$, so the definition is saying that $$\emptyset$$ and $$\mathbb{R}$$ are open sets in $$F$$ -- as must be the case with every topology on $$\mathbb{R}$$.
The complement of $$(0,1)$$ is $$(-\infty, 0] \cup [1, \infty)$$. This is not an open set in the topology.
The closure $$\bar{X}$$ of $$X \subset \mathbb{R}$$ is the minimal closed set containing $$X$$.
The interior of $$X$$ is the maximal open set contained in $$X$$.