TITLE: Is the cardinality of an open set in a complete metric space $X$ with no isolated points equal to $|X|$?
QUESTION [6 upvotes]: Let $(X,d)$ be a complete metric space with no isolated points.

Is it true that any open set in $X$ (in the metric topology) has the same cardinality as $X$?

It's true if $X = \mathbb{R}$, for example, so the general result seems somewhat plausible. 
I know that the Baire category theorem implies that $X$ is uncountable and also that any open set in $X$ is second category. So if $|X| = \mathfrak{c}$ (the cardinality of the continuum), then, assuming the continuum hypothesis, the cardinality of any open set $O$ is also $\mathfrak{c}$: if $O = \{x_1,x_2,... \}$ were countable, then it would be first category since each $x_i$ is nowhere dense in the metric topology.
What can we say when $|X| > \mathfrak{c}$?

REPLY [2 votes]: Seeing as the topological sum of two completely metrizable spaces is completely metrizable, your question is tantamount to asking whether all complete metric spaces without isolated points have the same cardinality, or in other words, whether every complete metric space without isolated points has the cardinality of the continuum.
The answer is no, because non-separable Hilbert spaces can have arbitrarily large cardinalities.