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

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    $\begingroup$ $\varnothing$ 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? $\endgroup$
    – Randall
    Sep 26, 2023 at 17:07
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    $\begingroup$ So it's both closed and open at the same time ? $\endgroup$ Sep 26, 2023 at 17:08
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    $\begingroup$ Also, it's not $\{\varnothing\}$, but just $\varnothing$. $\endgroup$
    – Randall
    Sep 26, 2023 at 17:08
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    $\begingroup$ Yes, just like $\varnothing$ is both open and closed. $\endgroup$
    – Randall
    Sep 26, 2023 at 17:08
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    $\begingroup$ 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. $\endgroup$ Sep 26, 2023 at 17:11

2 Answers 2

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$\Bbb R\setminus\Bbb R=\{\emptyset\}$ 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.

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    $\begingroup$ 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. $\endgroup$
    – MJD
    Sep 26, 2023 at 17:23
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    $\begingroup$ @MJD That's fair. But imo the best definition of closed is this one; any topology student should be aware of it. $\endgroup$
    – FShrike
    Sep 26, 2023 at 17:25
  • $\begingroup$ 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 $\endgroup$ Sep 26, 2023 at 18:48
  • $\begingroup$ @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. $\endgroup$
    – FShrike
    Sep 26, 2023 at 18:56
  • $\begingroup$ @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) $\endgroup$ Sep 26, 2023 at 19:02
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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 $(0,1]$.

To answer your question:

  • 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.

Which I guess isn't an answer to your question at all, since the answer is both yes and no.

Sorry, but that is the way it is.

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  • $\begingroup$ It was good of you to expand upon this point. +1 $\endgroup$
    – FShrike
    Sep 26, 2023 at 17:27
  • $\begingroup$ Thanks, I thought your answer was better because it hit all the important issues more succinctly! $\endgroup$
    – MJD
    Sep 26, 2023 at 17:29

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