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Consider the interval $X= [0,1)$. Is there a topology which makes the interval Hausdorff and compact?

My intuition tells me that such a topology cannot be found. I have attempted to prove that, if $\{a_n\}_n$ is a sequence which approaches $1$ from within the interval, $X\setminus a(\mathbb{N})$ would have to be open under a Hausdorff topology. If that were the case, one may be able to find $W_n$ open which such that $a_k \not \in W_n, \forall k >n$, and then $\{X\setminus a(\mathbb{N})\} \cup \{a_k\}_{k \in \mathbb{N}}$ would be an open cover for $X$ with no finite subcover, proving that the set cannot be compact.

There are many holes in my argument, though. For instance, I have been unable to prove the closedness of $a(\mathbb{N})$. Am I on the right track? Any help is appreciated.

francescoriccardocrescenzi
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  • Such a space would have to be normal as well. – ncmathsadist Jun 13 '23 at 13:32
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    You are using some unstated assumptions about how your proposed topology interacts with the order structure. Otherwise: the half open interval has the same cardinality as the closed interval, so transfer the usual topology with a (discontinuous) bijection. – Ethan Bolker Jun 13 '23 at 13:35

2 Answers2

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There exists a bijection $b:[0,1]\to [0,1)$.

Now, define the topology on $[0,1)$ as $$\mathcal T_{[0,1)} = \{b^{-1}(X)| X\text{is an open set in the standard Euclidean topology on }[0,1]\}.$$

It is trivial to show that $b$ is a homeomorphism between $([0,1), \mathcal T_{[0,1)})$ and $[0,1]$, equipped with the standard Euclidean topology.

Then, $([0,1), \mathcal T_{[0,1)})$ is a compact Hausdorff topological space, because it is homeomorphic to $[0,1]$.

5xum
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  • Thank you. So, if I understand correctly, given a topological space $(X,T)$ and a set $Y$ with the same cardinality as $X$, it is always possible to use this construction to find a topology $S$ on $Y$ such that $(X,T)$ and $(Y,S)$ share every property which is an invariant of homeomorphisms. – francescoriccardocrescenzi Jun 13 '23 at 13:54
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    @francescoriccardocrescenzi Precisely. This actually extends even beyond topology, by the way. Any type of structure you have on a set can be extended to another set via bijections. So, if $X$ is a vector space, and $Y$ is a set of equal cardinality, then you can construct addition and scalar multiplication on $Y$ such that $Y$ is a vector set, isomorphic to $X$. Or if $X$ is a group, then we can define an operation on $Y$ such that $Y$ is a group, isomorphic to $X$. Or if $X$ is a partially ordered set, then we can define an order on $Y$ such that $X$ is order isomorphic to $Y$. And so on. – 5xum Jun 13 '23 at 13:57
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The answer 5xum has given is correct and handles the general case very well. I think in this special case, one can construct a more 'obvious' example, which I would like to add. Namely notice that the map $f : [0,1) \to \mathbb C$, $f(t) = \mathrm{e}^{2i \pi t}$ defines a bijection from $[0,1)$ onto the unit circle $S^1$. We may now define a topology $\mathcal T$ on $[0,1)$ by the pullback under $f$, i.e. the open sets are inverse images of open sets in $S^1$. As $S^1$ is compact, so is $([0,1), \mathcal T)$.

Intuitively, this map wraps $[0,1)$ around in such a way, that the 'missing 1' attaches onto $0$, hence there are no more 'missing points' and our topology is compact.

Sven-Ole Behrend
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    To add to this, the basis of this topology would be ${(a, b)| a, b\in\mathbb R\land a<b}\cup {[0,a)\cup (b, 0)| a,b\in \mathbb R\land a < b}$ – 5xum Jun 13 '23 at 14:05