# Large discrete subsets of connected $T_2$-spaces

If $(X,\tau)$ is a topological space, we say $S\subseteq X$ is discrete, if the subspace topology on $S$ inherited from $(X,\tau)$ is discrete.

Let $\kappa$ be an infinite cardinal. Is there a connected $T_2$-space $(X,\tau)$ and a discrete subset $S\subseteq X$ such that $|X| = |S| = \kappa$?

• Assume that $H$ is a Hilbert space of dimension $\kappa$. That is the cardinality of a maximal orthonormal set is $\kappa$. Let $B$ be this Hilbert space base. Am I mistaken to think that the union of all finite dimensional subspace spaned by elements of $B$ is a possible candidate for your question, if $\kappa$ is not countable? – Ali Taghavi Jan 6 '17 at 14:43
• @AliTaghavi: That works for $\kappa \ge \mathfrak{c}$, of course. And in fact, in that case, you can just take $X=H$, since $H$ itself also has cardinality $\kappa$ (exercise). – Nate Eldredge Sep 9 '18 at 19:25

For any cardinal $\kappa$ at least the size of the continuum, the "really long line" of length $\kappa$ is an example.
This space, let's call it $L_\kappa$, is defined as follows. Begin with the ordinal $\kappa$ with its usual order topology. Then, for any $\alpha \in \kappa$, connect $\alpha$ and $\alpha+1$ with a copy of the unit interval.
$L_\kappa$ obviously has cardinality $\kappa$, and a discrete subspace of size $\kappa$ is given by the set of all successor ordinals.
If $\kappa < \mathfrak{c}$, then you can get an example by modifying the construction of $L_\kappa$: simply replace the unit interval with a countable connected Hausdorff space. (It's not obvious, but countable connected Hausdorff spaces do exist.)
If you want a closed discrete set take the metric hedgehog with $\kappa$ many spines, where $\kappa$ is the desired cardinality. If $\kappa<\mathfrak{c}$ replace $[0,1]$ by a countable connected Hausdorff space, as in Will Brian's answer.