4

It is well known that the interval $(0,1)$ is homeomorphic to $(0,+\infty)$. However, considering their closures, we have:

  • The closure of $(0,1)$ is $[0,1]$, which is compact.
  • The closure of $(0,+\infty)$ is $[0,+\infty)$, which is not compact.

I am confused because homeomorphism implies that topological properties should be preserved, and compactness is a topological property. How can this discrepancy be explained?

Harikrishnan S
  • 81
  • 1
    Homeomorphism preserves the topological properties of a space, not of its closure as a subspace of an arbitrary larger topological space. $(0,1)$ (and similarly $(0,\infty)$) can have various (non-homeomorphic) closures, in various spaces of which it is a subspace. For instance the closure of $(0,1)$ in $S^1$ (via $t\mapsto e^{2i\pi t}$) is $S^1$. – Anne Bauval Jun 06 '24 at 12:14
  • 1
    Possibly useful here: https://math.stackexchange.com/a/4855136/25554 which begins “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 $$”. – MJD Jun 06 '24 at 12:47

3 Answers3

4

As spinosarus123's answer already explains, closure is something that applies to subspaces of some given space, in this case $\mathbb{R}$. As such, there is no reason to expect a homeomorphism from $A$ to $B$ to extend to a homeomorphism (or even a map) between their closures as subspaces of some space, since this depends on the ambient space. There is, however, a situation in which homeomorphisms preserve closure, namely if you're starting out with a homeomorphism of the ambient space that restricts to a homeomorphism on $A$ and $B$.

In your particular case, you should therefore only expect compatibility with closures if you're given a homeomorphism $\mathbb{R} \to \mathbb{R}$ to begin with, and there's no such homeomorphism mapping $(0, 1)$ to $(0, \infty)$.

Ben Steffan
  • 8,325
1

As the example in your post amply demonstrates, using two different subsets of the topological space $X = \mathbb R$, the "closure of $A$ in $X$" is not a topological property of $A$: it also depends on $X$.

What one can say in this situation is that the closure is a topological property of a space-and-a-subset. That is, given a space $X$ and a subset $A$, and given another space $B$ and a subset $Y$, if there exists a homeomorphism $h : X \to Y$ such that $h(A)=B$ then the closure of $A$ in $X$ is homeomorphic to the closure of $B$ in $Y$. So in that situation it would be true that the closure of $A$ in $X$ is compact if and only if the closure of $B$ in $Y$ is compact.

So for the example in your post, one can reach an interesting conclusion: there does not exist a homeomorphism $h : \mathbb R \to \mathbb R$ such that $h(0,1)=(0,\infty)$.

BCLC
  • 14,197
Lee Mosher
  • 135,265
  • Lee Mosher, indeed there is no such automorphism of $\mathbb R$ that maps either an open bounded interval to an open unbounded interval or a closed bounded interval to a closed unbounded interval right? – BCLC Jun 11 '24 at 21:42
  • 1
    That is correct. – Lee Mosher Jun 11 '24 at 22:26
  • 1
    But the point of my last sentence is that the nonexistence of such an automorphism is an immediate consequence of the fact that the closures of $(0,1)$ and $(0,\infty)$ in $\mathbb R$ are not homeomorphic. – Lee Mosher Jun 11 '24 at 22:28
0

The closure of these spaces are homeomorphic. The closure of the space $(0,1)$ is $(0,1)$ and the closure of the space $(0,\infty)$ is $(0,\infty)$.

The thing you are referring to as closure is the closure of the set $(0,1)$ as a subset of the space $\mathbb{R}$. This closure is indeed $[0,1]$, however this is not intrinsic to the space $(0,1)$.

spinosarus123
  • 2,612
  • 1
  • 8
  • 22
  • Perhaps it'd be more careful to say that the closure of a space is simply undefined, although we can consider the closure of the entire set as a subspace (which is always itself). – Steven Clontz Jun 12 '24 at 00:37