Is $\left|\left[\frac{1}{7,000,000,000},1\right]\right|<\left|[0,1]\right|$? My question concerns the cardinality of two subsets of $\mathbb{R}$. It feels very naive, but, isn't $\left[\dfrac{1}{7,000,000,000},1\right]$ "smaller" than $[0,1]$? I think that this is dependent upon one's set theory axioms, but I'd like an answer from the most common axiomatic models.
After consulting Wikipedia, I've seen that you must show there is only an injective function from $B$ to $A$ to prove that $|A|>|B|$. Such a function is obvious: $f:A\to B$ with $x\mapsto x$. The hard part seems showing there is not a bijective function.
I don't know how to do that, but I think I might be able to figure out why I don't know how to do that. It seems like the continuum hypothesis is at work here: I am trying to find some sort of cardinality, $X$, such that $\aleph_0 < |X| < 2^{\aleph_0}$. This is because it would be intuitive that any subset of the real number system which is an interval would have a larger cardinality than $\aleph_0$. Similarly, I'm trying to find a similar cardinality $Y$ such that $X<Y$.
What's going on here?
 A: All intervals have the same cardinality as the reals $2^{\aleph_0}$.  A Blumenthal has given you a bijection between any closed interval and $[0,1]$.  Then you can find a bijection between $[0,1]$ and $(0,1)$ (how can two points matter?) and between $(0,1)$ and $\mathbb R$.  It is true that the length of the intervals is different, but the cardinality is not.
According to the continuum hypothesis, known to be consistent with ZFC but not required, there are no sets $X$ such that $\aleph_0 < |X| < 2^{\aleph_0}$, so don't look too hard for them.
A: Just to supplement Ross’ already-very-clear answer. In what follows, I shall assume that you know some basic measure theory.
Let $ \mu $ be the standard Borel measure defined on the Borel $ \sigma $-algebra $ \mathcal{B} $ of the closed interval $ [0,1] $. Define a quasi-ordering $ \preceq $ on $ \mathcal{B} $ as follows:
$$
\forall B_{1},B_{2} \in \mathcal{B}: \quad B_{1} \preceq B_{2} \stackrel{\text{def}}{\iff} \mu(B_{1}) \leq \mu(B_{2}).
$$
If $ B_{1} $ and $ B_{2} $ are sub-intervals of $ [0,1] $, then $ \preceq $ is simply a comparison of their lengths.
Define also a quasi-ordering $ \leqslant $ on $ \mathcal{B} $ as follows:
\begin{align}
\forall B_{1},B_{2} \in \mathcal{B}: \quad
                             B_{1} \leqslant B_{2}
&\stackrel{\text{def}}{\iff} |B_{1}| \leq |B_{2}| \\
&\iff                        (\exists f)(\text{$ f: B_{1} \to B_{2} $ is injective}).
\end{align}
The point to note is:

The quasi-ordered set $ (\mathcal{B},\preceq) $ is not isomorphic to the quasi-ordered set $ (\mathcal{B},\leqslant) $ in the category of quasi-ordered sets.

