# Cardinality of an arbitrary interval of real numbers

Let $$a, b\in\mathbb{R}$$ with $$a. Prove that $$|\{x\in \mathbb{R}\ | \ a< x< b\}|=|\{x\in \mathbb{R}\ | \ 0.

Would constructing a bijection be the most effective way to prove this? Or should I consider proving that the real numbers from a to b are uncountable?

• You want to create a bijection. This would be easiest. – D.B. Apr 8 at 22:40
• Producing a bijection shouldn't be hard. Think of the equation of a line. – ZeroXLR Apr 8 at 22:41
• Think about $y=\frac1{b-a} (x-a)$ – Peter Foreman Apr 8 at 22:44

You want to map $$(a,b)$$ onto $$(0,1)$$. The easiest way is to consider the equation of the line through $$(0,a)$$ and $$(1,b)$$. This line has equation $$y=(b-a)x+a$$. So the map $$f:x\mapsto (b-a)x+a$$ is bijective from $$(0.1)$$ onto $$(a,b)$$. So the two intervals have the same cardinality.

• Yes. Completely correct. – Jethro Apr 8 at 23:25

Just showing that $$(a,b)$$ is uncountable wouldn't actually solve the problem - it's consistent with the usual axioms (ZFC) of set theory that there are uncountable sets of reals strictly smaller in cardinality than $$(0,1)$$.

Instead, you need to whip up a bijection. But this won't be too hard. As a first step, maybe consider how to get a bijection between $$(0,1)$$ and $$(0,b)$$ for $$b>0$$; next, given an interval $$(a,b)$$, is there some easy way to put it into bijection with an interval of the form $$(0,c)$$?
Now let me give a bit more detail about the claim I made at the beginning of this answer. The statement that every uncountable set of reals has the same cardinality as $$(0,1)$$ is (one way to phrase) the continuum hypothesis (CH). CH, like many other reasonable-sounding statements about infinite cardinalities, is neither provable nor disprovable from ZFC; that's not to say that no proof or disproof is currently known, but rather that we can prove that ZFC can neither prove nor disprove CH.$$^1$$ This was established by Godel and Cohen - Godel showed that ZFC can't disprove CH, and Cohen showed that ZFC can't prove CH. Both methods of proof became fundamental to modern set theory.
$$^1$$Unless ZFC is inconsistent, in which case it proves and disproves everything.