Question

Assume that $X,Y$ are infinite dimensional Banach spaces. Is it true that if density character of $X$ is less then or equal to density character of $Y$ then $card X \leq card (Y)$ ?

Answer 1

Let $\aleph$ be the density character of the Banach space $X$ and let's compute the cardinality of $X$ using AC but not much set theory, since these days cardinal arithmetic is generally given short shrift in real analysis courses.

Take a dense set $D=\{x_a: a<\aleph\}$ in $X$ and observe that all tails of this set are still dense, so for each $x$ in $X$ there is a sequence $x_{a_0}<x_{a_1} <\dots$ in $D$ that converges to $x$. This gives an upper bound for the cardinality of $X$.

To get the corresponding lower bound, take a set $\{y_a: a<\aleph\}$ of unit vectors in $D$ so that for each $a < \aleph$, the distance of $y_a$ to the span of $\{y_b: b<a\}$ is larger than $1/2$. Given $A=\{a_0<a_1<\dots<\aleph\}$ define $z_A = \sum_{n=0}^\infty 10^{-n} y_{a_n}$. It remains to show that if $A\not=B$ then $z_A\not= z_B$.

Write $A=\{a_0<a_1<\dots<\aleph\}$ and $B=\{b_0<b_1<\dots<\aleph\}$. If $a_0\not= b_0$, then $\|y_{a_0}-y_{b_0}\|> 1/2$ and the desired result follows from the triangle inequality. In the general case let $m$ be the smallest $n$ s.t. $y_{a_n}\not= y_{b_n}$. Apply the first case to $10^m\sum_{n=m}^\infty 10^{-n} y_{a_n}$ and $10^m\sum_{n=m}^\infty 10^{-n} y_{b_n}$.

Answer 2

Answering your comment: This is not true for metric spaces in general.
The discrete space of size $\aleph_1$ has density $\aleph_1$ and is metric (two distinct points have distance 1, this metric even gives a complete metric space), but the real line has density $\aleph_0$ and size $2^{\aleph_0}$, which can be larger than $\aleph_1$.