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In the book "Introduction to Topology" by Bert Mendelson, ch3, theorem 4.7 states that for any Topological Space (X, T) $\bar{X} = X$.

Now I understand that any subset of $X$ is also a subset of $\bar{X}$, but don't see why $X = \bar{X}$.

An answer with examples would be appreciated

Cedric Martens
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  • To the person who asked what definition of closure I was using (comment was deleted). Let $A$ be a subset of a topological space. A point $x$ is said to be in the closure of $A$ if, for each neighborhood $N$ of $x$, $N \cap A \neq 0$. The closure of $A$ is denoted $\bar{A}$ – Cedric Martens Feb 01 '24 at 18:59
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    Don’t leave clarifications in comments, EDIT THE QUESTION. – user3840170 Feb 02 '24 at 10:37
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    @user3840170 You are right, but caps is unnecessary. – preferred_anon Feb 02 '24 at 11:27
  • One can also edit the definition into the question oneself. https://en.wikipedia.org/w/index.php?title=Wikipedia:JUSTDOIT&redirect=no – MJD Feb 02 '24 at 14:27

2 Answers2

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Your confusion is a common one. You haven't realized that "closure" is a “relative” concept. This means that it's not quite correct to talk about ‘the’ closure of a set $A$. Rather, $A$ has a closure when considered as a subset of some larger space $X$, and the closure of $A$ might be different depending on what $X$ is.

This is similar to how the idea of a set complement is relative. What's the complement of the set of even integers? Well, if you consider it to be a subset of $\Bbb Z$, then its complement is the set of odd integers. But if you consider it to be a subset of $\Bbb R$, the complement includes not just the odd integers but also non-integers like $\frac12$.

Context matters, and when we say "the complement of $A$” we depend on the listener to recognize the context. Similarly when we say “the closure of $A$”.

The notation $\bar A$ is misleading, because what $\bar A$ is will depend on $X$, but the $X$ doesn't appear in the notation.

Here's Mendelson's definition of “closure” (Definition 4.2, page 124):

Let $A$ be a subset of a topological space. A point $x$ is said to be in the closure of $A$ if, for each neighborhood $N$ of $x$, $N\cap A\ne \emptyset$.

This depends on what the neighborhoods of the enclosing space are. Mendelson has obscured this by not even naming the enclosing space, which I have called $X$. He implies, but doesn't say, that the points $x$ are points of the space $X$.

I'm going to rephrase Mendelson's definition to make this explicit:

The closure of a subset $A$ of a topological space $X$ is the set of all points $x\in X$ such that, for every open set $N$ of $X$ containing $x$, $$N\cap A\ne\emptyset.$$

Now let's do the example you asked about. Let $A=(0,1)$ and $X =\Bbb R$ with the usual topology. The closure of $A$ in $\Bbb R$ is $[0,1]$ as I think you know.

But now let $A=(0,1)$ and $X = \Bbb R^+ = (0, \infty)$, again with the usual topology. The closure of $A$ in $\Bbb R^+$ is $(0, 1]$.

Why is the point $0$ missing from $\bar A$? Because the definition says

The closure… is the set of all points $x\in X$ such that…

and if $X=\Bbb R^+$, then $0\notin X$, so $0$ can't be in the closure of $A$. It's not in the closure because it's not part of the universe!

Similarly, you can consider $X = (0,1)$ with the usual topology. The closure of $A=(0,1)$ in this space is $(0,1)$. It doesn't include points $0$ or $1$ because those are not part of the space at all.

And this is the example you're asking about: If $(0,1)$ is the whole space, then its closure is $(0,1)$.

MJD
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Closure is an operator that takes subsets of $X$ to subsets of $X$ by definition (regardless of which one you use). So of course $\overline{X}$ has to be contained in $X$. Closure of any subset of $X$ is contained in $X$. On the other hand $X\subseteq\overline{X}$ by another property of closure. So they have to be equal, no other choice.

freakish
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  • So that means the set $X$ is closed for every Topological Space ? What if $X$ is the open interval (2, 5). I'm confused here – Cedric Martens Feb 01 '24 at 19:02
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    @CedricMartens yes, $X$ is closed in itself. Being "closed" is always relative to some other space. The $(2,5)$ interval is closed in $(2,5)$, not in $\mathbb{R}$. – freakish Feb 01 '24 at 19:06
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    @CedricMartens because you cannot take a point outside of itself. If you look at $X$ then nothing else exists. If you look at $(2,5)$ interval as a standalone topological space, then nothing else exists at that moment. – freakish Feb 01 '24 at 19:14
  • okay I see it, yes for each neighborhood of any x in (2,5), the intersection of the neighborhood and (2,5) is non empty, therefore closed thank you! – Cedric Martens Feb 01 '24 at 19:14
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    @CedricMartens no, that's not a correct reasoning. This reasoning is valid for any subset. It pretty much means $A\subseteq\overline{A}$. The correct reasoning is: I cannot take a point outside of $(2,5)$ to even test if it belongs to closure of $(2,5)$. Not if $(2,5)$ is my universe at the moment. I could do it for $\mathbb{R}$, that's why inside it closure of $(2,5)$ is strictly bigger. But not in $(2,5)$ universe. – freakish Feb 01 '24 at 19:19
  • @CedricMartens A tip: Never say "The set X is closed", or even worse "The space X is closed"; say instead "The set X is a closed subset of Y". This prevents the confusion from ever possibly arising. – preferred_anon Feb 02 '24 at 11:31