Let X be a set. Is it possible for X to have an infinite number of topologies up to homeomorphism (i.e. infinitely-many different topological structures)?
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2Every manifold of dimension $\ge 1$ has the same cardinality as the real numbers. So that is a whole lot of different topologies right there. – Nate Nov 24 '14 at 21:51
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1In fact a set of infinite cardinality $\kappa$ has $2^{2^\kappa}$ non-homeomorphic topologies: see http://math.stackexchange.com/questions/65731/what-is-the-cardinality-of-the-set-of-all-topologies-on-mathbbr – Robert Israel Nov 24 '14 at 22:00
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possible duplicate of Are there uncountably many non homeomorphic ways to topologize a countably infinite set? – Nate Eldredge Nov 25 '14 at 03:05
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Yes. Take for example the real line plus some finite number of isolated points. You can change the topology so that any number of the points are isolated and still have uncountably many left over, and no two of these topologies are homeomorphic.
hardmath
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MartianInvader
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@user48481MirkoSwirko Whoops, you are correct, edited the answer. – MartianInvader Nov 24 '14 at 21:46
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Yes. For example, we can take $\Bbb N$ under the topologies defined as follows: $$ \tau_0 = \{\emptyset,\Bbb N\}\\ \tau_1 = \tau_0 \cup \{1\} \cup (\Bbb N \setminus \{1\})\\ \vdots\\ \tau_n = \{\emptyset,\Bbb N\} \cup\mathcal P(\{1,\dots,n\}) \cup \{(\Bbb N \setminus S): S \in \mathcal P(\{1,\dots,n\})\} $$ each of these topologies consist of different finite numbers of open sets, so no two are homeomorphic.
Ben Grossmann
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Let $\mathcal S=\{X\subseteq\mathbb R:|X|=\mathfrak c\}$ where $\mathfrak c=2^{\aleph_0}$. For each $X\in\mathcal S$, there are only $\mathfrak c$ continuous functions $f:X\to\mathbb R$ (since such a function is determined by its restriction to a countable dense subset of $X$); hence $|\{Y\in\mathcal S:Y\text{ is homeomorphic to }X\}|\le\mathfrak c$. Since $|\mathcal S|=2^\mathfrak c$, it follows that there are $2^\mathfrak c$ non-homeomorphic sets in $\mathcal S$; in other words, $\mathbb R$ has $2^\mathfrak c$ non-homeomorphic subspaces of cardinality $\mathfrak c$. Since each of those subspaces is homeomorphic to a topology on $\mathbb R$, there are $2^\mathfrak c$ non-homeomorphic separable metrizable topologies on $\mathbb R$.
bof
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