# Are infinities included in the closure of the real set $\overline{\mathbb{R}}$

I am facing something that I do not understand in topology.

My lesson states that for a given set E, any given topology of E must at least contain E and the empty set, and that these elements are called the opens. Furthermore a subset F is closed if E\F is an open.

I am struggling to understand what $$\overline{\mathbb{R}}$$ is. Indeed we have $$\mathbb{R} \setminus\mathbb{R} = \{\emptyset\}$$ an open, therefore $$\mathbb{R}$$ is closed (but it's an open as it's part of the topology actually), therefore we should have $$\overline{\mathbb{R}} \neq \mathbb{R}$$ (since $$\overline{{E}} = {E}$$ if and only if E is closed).

Perhaps I thought that $$\overline{\mathbb{R}}$$ was $$[-\infty ,\infty]$$, but it seems that this is not the case, so I am kinda lost there.

Is $$\overline{\mathbb{R}} = \mathbb{R}$$ ?

• $\varnothing$ is an open set, so $\mathbb{R}$ is closed. So, it's closure is itself. Why do you think that $\mathbb{R} \setminus \mathbb{R}$ being open means it is not closed? Sep 26, 2023 at 17:07
• So it's both closed and open at the same time ? Sep 26, 2023 at 17:08
• Also, it's not $\{\varnothing\}$, but just $\varnothing$. Sep 26, 2023 at 17:08
• Yes, just like $\varnothing$ is both open and closed. Sep 26, 2023 at 17:08
• In general, the closure of a subset $A$ of a topological space $X$ is still a subset of $X$. The closure can never introduce new points to the space. Sep 26, 2023 at 17:11

$$\Bbb R\setminus\Bbb R=\{\emptyset\}$$ an open, therefore $$\Bbb R$$ is not closed

This is wrong on $$2$$, perhaps $$3$$ points. Firstly, $$\Bbb R\setminus\Bbb R=\emptyset$$ which is not the same thing as $$\{\emptyset\}$$. Secondly, if $$\Bbb R\setminus K$$ is open, then $$K$$ must be closed... by definition of "closed". So actually you've shown $$\Bbb R$$ is closed. Thirdly, I think you're making the common beginner mistake that "not closed" and "open" mean the same thing - this is absolutely false. In any topological space $$X$$, $$X$$ is both open and closed ("clopen") (similarly for $$\emptyset$$) and if your space is disconnected then there'll be some more examples of these "clopen" sets.

The closure of $$\Bbb R$$ in itself is $$\Bbb R$$; that's true of any space. However $$\overline{\Bbb R}$$ usually denotes something special, the extended reals $$\Bbb R\cup\{-\infty,+\infty\}$$ (where $$\pm\infty$$ are two 'formally adjoined' points, they have no meaning, they are not "infinite numbers", they are just some distinct sets which aren't elements of $$\Bbb R$$) which carries a certain topology. This isn't really a closure operation; however, it is a compactification of $$\Bbb R$$, and it is true that the closure of $$\Bbb R$$ in $$\overline{\Bbb R}$$ is $$\overline{\Bbb R}$$, i.e. $$\Bbb R$$ is a dense subspace of the extended reals.

• In some presentations, a closed set is defined as one that contains its limit points, and then it is a theorem, not a definition, that closed sets are complements of open ones.
– MJD
Sep 26, 2023 at 17:23
• @MJD That's fair. But imo the best definition of closed is this one; any topology student should be aware of it. Sep 26, 2023 at 17:25
• Thank you for your explanation, I have a question for the extended real numbers let $a \neq b$ with a,b positive reals, do we have $a\infty \neq b\infty$ (or for sums) or are all $\infty$ equal Sep 26, 2023 at 18:48
• @BloomeyPhysics This is no longer about topology, this is asking about algebra structure of $\overline{\Bbb R}$. Unfortunately there is no sensible ring (or field or …) structure for this space, e.g. what is $\infty+(-\infty)$? So I guess I would say, $a\infty$ is meaningless. But if you were going to define it, $a\infty=\infty=b\infty$ would be the best definition in my opinion. Sep 26, 2023 at 18:56
• @FShrike I see, this is because I am working on distances and I was wondering if $|\exp^{-x} - \exp^{-y}|$ would no longer be a distance if we had $2\infty>\infty$ (sorry I don't know why absolute values are not working) Sep 26, 2023 at 19:02

Closure is not an absolute notion. This is similar to the way that set complementation is not an absolute notion. Consider:

What is the complement of $$\Bbb R$$?

Anne might say that it was $$\emptyset$$, because $$\Bbb R$$ contains all real numbers. But Bob might say that it was $$\{ a+bi\mid b\ne 0\}$$ the set of all complex numbers with nonzero imaginary part. The correct answer depends on the context: Are you interested in $$\Bbb R$$ alone, as Anne is, or as a subset of the complex numbers, as Bob is?

The idea of “the complement of $$S$$” always depends on context, and in advanced mathematics one is more likely to see the notation $$X\setminus S$$ that makes clear what the context is: what set $$X$$ is the complement being taken “relative to”? In this notation Anne is thinking of $$\Bbb R\setminus \Bbb R=\emptyset$$ and Bob is thinking of $$\Bbb C\setminus \Bbb R$$.

Topological closures are the same way. It never makes sense to talk about the closure of a set without a clear idea of what the surrounding context is. When the context isn't completely clear, careful speakers avoid saying “the closure of $$S$$” and instead make the context explicit by saying “the closure of $$S$$ in $$X$$”.

Considered as a subset of $$\Bbb R$$, the closure of the open interval $$(0, 1)$$ is the closed interval $$[0, 1]$$.

But considered as a subset of $$(0,1)$$, the closure of $$(0,1)$$ is just $$(0,1)$$ again, because the closure of any whole space is just the same space.

And considered as a subset of $$(0,\infty)$$, the closure of $$(0,1)$$ is $$(0,1]$$.

• Considered as a subset of $$\Bbb R$$, the set $$\Bbb R$$ is already closed, and its closure is $$\Bbb R$$
• But considered as a subset of the larger space of “extended real numbers” $$[-\infty, \infty]$$, the closure of $$\Bbb R$$ is $$[-\infty, \infty]$$ and includes the endpoints.