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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.

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

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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 $x<y<0$.

No bounded set has any interior points so interior of $[0,1]$ is empty.

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$F$ 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 $X$.

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