Is the cardinality of an open set in a complete metric space $X$ with no isolated points equal to $|X|$?

Question

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 $|X| > \mathfrak{c}$?

Answer 1

It's not true.

Consider more-than-continuum-many copies of $[0,1]$ with the usual metric, "put together discretely" - that is, we set the distance between any two points in different copies to be $2$. This is complete, but open neighborhoods can have size continuum.

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Maybe a more natural example would be the "star-like" version of the above construction: fix some index set $I$ of arbitrary cardinality, and let $$M=(I\times (0, 1])\sqcup\{*\},$$ where we think of this as $I$-many copies of $[0, 1]$ glued together by identifying all of their $0$s (as $*$). This has a natural metric structure: just "follow the arms". It's clear what the distance between $(i, a)$ and $(i, b)$ should be for $i\in I$ and $a, b\in (0, 1]\cup \{*\}$; and if $a, b\in (0, 1]$ and $i,j$ are distinct elements of $I$, we'll set $d((i, a), (j, b))=a+b$.

With respect to this metric, $M$ is complete - but again, open neighborhoods can avoid the $*$ and live entirely in a single "arm," thereby having small cardinality.

In particular, note that this example is connected, unlike the previous one.

Answer 2

As others have pointed out, this is not true for $|X|>\mathfrak{c}$. However, it is true for $|X|\leq \mathfrak{c}$, even without assuming the continuum hypothesis. Here's a proof.

Let $U$ be a nonempty open set in a complete metric space $X$ with no isolated points. Let $B$ be some open ball whose closure $\overline{B}$ is contained in $U$. Then $\overline{B}$ is also a complete metric space with no isolated points, and it suffices to show $|\overline{B}|\geq \mathfrak{c}$. That is, we may as well just assume $U=X$, and show that any nonempty complete metric space $X$ with no isolated points has cardinality at least $\mathfrak{c}$.

To prove this, take two distinct points $x_0$ and $x_1$ in $X$ and open balls $U_0$ and $U_1$ of radius $\leq 1$ with disjoint closures containing them. Since $x_0$ is not isolated, $U_0$ contains two distinct points $x_{00}$ and $x_{01}$, and we can choose open balls $U_{00}$ and $U_{01}$ of radius $\leq 1/2$ with disjoint closures containing them and contained in $U_0$. Similarly, we can find two distinct points $x_{10}$ and $x_{11}$ contained in open balls $U_{10}$ and $U_{11}$ of radius $\leq 1/2$ with disjoint closures which are contained in $U_1$.

Continuing this process, we can find points $x_s$ and open balls $U_s$ for each finite sequence $s$ of $0$s and $1$s with the following properties. For each $s$, $x_s\in U_s$ and the radius of $U_s$ is at most $1/n$ where $n$ is the length of $s$. If $t$ is an initial segment of $s$, then $U_s\subseteq U_t$. For any $s$, the sets $U_{s0}$ and $U_{s1}$ have disjoint closures.

Now note that if $s$ is any infinite sequence of $0$s and $1$s, the points $(x_t)$ where $t$ ranges over the finite initial segments of $s$ form a Cauchy sequence (Cauchy because they are contained in balls $U_t$ whose radii go to $0$). By completeness, this sequence converges to some point we can call $x_s$. Note that different sequences give different points, since $x_s$ is in the closure of $U_t$ for each initial segment $t$ of $s$ and at each step we chose each new pair of balls to have disjoint closures. So this gives a distinct point of $X$ for each infinite sequence of $0$s and $1$s, proving that $|X|\geq\mathfrak{c}$.

Answer 3

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.