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Is real line an open or closed interval $(-\infty,+\infty)$? I found this written in my lecture slide.

"The entire real line is infinite interval that is both open and closed."

Can anyone explain what does that mean?

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    $\begingroup$ What is your definition of open and closed? (If you have the correct definition, you can start proving it). $\endgroup$ – user99914 Sep 30 '17 at 4:28
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First of all, note that closed set and closed interval are different things! Similarly open set and open intervals are different things.

For example: $\mathbb{N}$ is closed set but not a closed interval.

$(1,2) ∪ (3, 4)$ is open set in $\mathbb{R}$ but not an open interval.

Real line or set of real numbers $\mathbb{R}$ is both "open as well closed set". Note $\mathbb{R}$ not a closed interval, that is $\mathbb{R} ≠ [-∞, ∞]$.

For definition of open and closed sets, See here

https://en.m.wikipedia.org/wiki/Open_set

https://en.m.wikipedia.org/wiki/Closed_set

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$\mathbb R$ is the set of real numbers. Choose a topology on it, whatever which one. Then $\emptyset$ and $\mathbb R$ are open sets (because of axioms of a topology). Then $\mathbb R$ is not only open but closed because it's the complement of the open empty set. QED

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If you define open sets in $\mathbb R^n$ with a help of open balls then it can be proved that set is open if and only if its complement is closed. Then, it happens that any union (finite or infinite) of open sets is an open set and that any finite intersection of open sets is open.

But for this to work you have to take as convention that the empty set is open, because intersection of disjoint open sets is an empty set. Then the $\mathbb R^n$ is closed because it is complement of an empty set, which is open. But also, since union of infinite number of open sets can cover whole space $\mathbb R^n$ then, for above written to work again you have to take as convention that whole space $\mathbb R^n$ is open, but then, its complement, which is an empty set, is closed.

So you arrive at the conclusion that both empty set and $\mathbb R^n$ are sets that are both open and closed. I hope that you went through all this terms here, so that you can understand why it is as it is.

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Per definitions in the abstract topological setting, the adjective "closed" is capturing the fact that the interval contains all of its limit points$^\dagger$, whereas "open" is capturing the fact that, around every $x \in I$, there will exist a $\delta, \varepsilon \in \mathbb{R}$ such that we'll have a sub-interval $(x\!-\!\delta, d\! +\! \varepsilon) \subset I$.

Both of these things are indeed true. The fact that $(-\infty, \infty)$ is open is "intuitively" true (I'll leave a formal proof to you), and the fact that $(-\infty, \infty)$ is closed is guaranteed by the fact that Cauchy sequences in $\mathbb{R}$ converge.


$^\dagger$ This definition of "closed" is more commonly seen in analysis with respect to metric spaces. Metric spaces are a special type of topological space; so more generally, if $X$ is a topological space, we call a subset $U \subset X$ closed if and only if the set $X \setminus U$ is open. The notion of a "limit point" can be generalized to arbitrary topological spaces, and from here it can be shown that the two definitions of "closed" are equivalent.

In our scenario, the complement of $\mathbb{R}$ is $\emptyset$, which is open by definition. Or, thinking analytically, it is vacuously true that $\emptyset$ contains all its limit points.

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  • $\begingroup$ Did you mean there will exist $\delta\in\mathbb{R}$ such that $\delta>0$ and $(x-\delta,x+\delta)\subset I$? $\endgroup$ – Rosie F Sep 30 '17 at 8:46
  • $\begingroup$ Yes I did. Thanks for catching that! $\endgroup$ – Kaj Hansen Sep 30 '17 at 16:12
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It's open and closed in the topological sense (see here). We say $A \subseteq \mathbb{R}$ is open if, for every $a \in A$, there is $r >0$ small enough such that $(a-r, a+r) \subseteq A$. Ok, we can equip the real line with different topologies, but it's the standard one. In this sense, $\mathbb{R}$ is a open set. Nice. On the other hand, $\varnothing = \mathbb{R} \setminus \mathbb{R}$ is open too, then $\mathbb{R}$ is closed (it's the definition of closed set). And, indeed, $\mathbb{R}$ is an interval (it's connected).

Note that all I said has no relation with sup/maximum or inf/minimum: $\mathbb{R}$ is diffeomorphic to $(0,1)$, it has no max/min. So, I think it isn't a good ideia say $\mathbb{R}$ is a closed interval (it's a interval, a closed set, but we can't say it has max/min).

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