Can a Set Have Infinitely-Many Non-Homeomorphic Topologies? 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)?
 A: 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.
A: 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.
A: 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 $\mathbb R$.
