\(X\) is a topological space with infinite cardinality which is homeomorphic to \(X \times X\). Then

A. \(X\) is not connected.

B. \(X\) is not compact.

C. \(X\) is not homeomorphic to a subset of \(\mathbb{R}\)

D. none of the above


Let us take \(X=\mathbb{Z}\). The topology is the subspace topology obtained from \(\mathbb{R}\). Then, in fact, \(X\) has the discrete topology. This is because for every \(n\in \mathbb{Z}\), \((n-1,n+1)\) is an open subset of \(\mathbb{R}\) and hence \((n-1,n+1)\cap \mathbb{Z}=\{n\} \) is open in \(\mathbb{Z}\).

We have a bijection from \(\mathbb{Z} \times \mathbb{Z}\) to \(\mathbb{Z}\). The product of two discrete spaces is again a discrete space because we have all sets of the form \(\{n\} \times \{m\} =\{(n,m)\} \) as open sets. Now, any map from a discrete space is continuous. Because the inverse image of any open set is a subset of the space and every subset is open in that space. So in fact, the bijection that we know of from \(\mathbb{Z} \times \mathbb{Z}\) to \(\mathbb{Z}\) is continuous in both ways. Therefore in this setup \(X \times X\) is homeomorphic to \(X\). And also \(X=\mathbb{Z}\) is a subset of \(\mathbb{R}\). So this provides us with a counterexample for option C.

Next, we will take \(X=\mathbb{Z}\) to disprove A and B together. But now we will consider a different topology for \(\mathbb{Z}\). We will consider the indiscrete topology. That is, the only sets which are open in \(X\) are \(X\) and the empty set. Note firstly, that since we only have finitely many open sets to start with, \(X\) is compact. \(X\) is also connected because the only non-empty set which is both open and closed is \(X\) itself. We use the same bijection that we know of between \(\mathbb{Z}\) and \(\mathbb{Z}\times \mathbb{Z}\). Note that if a map has its co-domain set as an indiscrete space then that map must be continuous. This is because the only open sets are empty and whole set, and their inverse images are empty and the whole domain set respectively which are always open. Also, note that if \(X\) has the indiscrete topology then so does \(X \times X\). Because the open sets in \(X \times X\) are \(\phi \times \phi = \phi , \phi \times X = \phi , X \times \phi = \phi, X \times X \). So both the spaces in consideration have indiscrete topology. So the bijection that we have is continuous in both ways. This, therefore, gives the counterexample for A and B.