Build a topological manifold starting from a set. Suppose you are given a generic set $X$. There exist sufficient and non-trivial conditions that ensure the existence of a topology $\tau_X$ on X such that the topological space $(X,\tau_X)$ is a topological manifold?
 A: Yes, with restriction only on cardinality - depending on your definition of manifold. For me, a manifold is a second countable Hausdorff space locally homeomorphic to a Euclidean space. Let's consider the various cases involved here.
In dimension $0$, any manifold is discrete. Second-countable implies that $|X|$ is countable in this case, as any base for the topology of a discrete space must contain the singletons. Given any finite or countable set, giving it the discrete topology makes it a manifold, so all sets up to countable support a topology that makes that set a manifold.
In dimension $n>0$, our cardinality is at least $|\Bbb R|$ (as we have a neighborhood homeomorphic to $\Bbb R^n$, which is cardinality $|\Bbb R|$. But any (second-countable) manifold has to have cardinality at most $|\Bbb R|$, as follows. The topology itself must be at most cardinality $|\Bbb R|$; for every element of $\tau_X$ can be written as a union of elements of the countable basis. Picking some such decomposition for every open set gives us an injective map $\tau_X \to \mathcal P(\Bbb N)$. But there's an injective map $\Bbb R \to \tau_X$ given by picking some injection $f: \Bbb R \to X$ and writing $f': \Bbb R \to \tau_X$ by $f'(x) = \{f(x)\}^c$ (the complement of the point $f(x)$). So by Cantor-Schroder-Bernstein (and recalling that $|\Bbb R| = |\mathcal P(\Bbb N)|$), we see that $|\tau_X| = |\Bbb R|$, and considering the injective map $f: \Bbb R \to X$ and the canonical injection $X \to \tau_X$ we see that $|X| = |\Bbb R|$. (This proof actually shows that any T1 space has cardinality at most $|\Bbb R|$; we used the manifold property to show that a positive-dimensional manifold has cardinality equal to $|\Bbb R|$.)
So if your manifolds are second-countable, the condition is precisely: $X$ is finite, countable, or has cardinality equal to $|\Bbb R|$. But if they're not second-countable, any set supports the topology of a manifold: just take the discrete topology $(X, \mathcal P(X))$! Even if you require that your manifolds are paracompact one easily sees that any discrete space is paracompact. So the conditions on $X$ are only on cardinality, no matter your definition. 
Edit: Of course, we have to show that any set $X$ with $|X| = |\Bbb R|$ supports a manifold structure. Pick a bijection $f: \Bbb R \to X$; this induces a bijection $f^*: \mathcal P(\mathcal P(\Bbb R)) \to \mathcal P(\mathcal P(X))$. Let $\tau$ be $\Bbb R$ with the standard topology, and consider $X$ with the topology $f^*(\tau)$. Then it's easy to see that $f$ defines a homeomorphism between $(\Bbb R, \tau)$ and $(X, f^*(\tau))$, showing that $X$ supports a manifold structure.
