# Prove that a half-open line is uncountable - corrected

I have to provide a proof that a halfopen line $$L_a:={𝑥 ∈ ℝ:x > a}$$ is uncountable.

I have used Schröder-Bernstein-Cantor and tried to show that an injection exists between $$L_a$$ and $$ℝ$$ an injection between $$ℝ$$ and $$L_a$$ to conclude that a bijective function exists between $$ℝ$$ and $$L_a$$ to prove that $$L_a$$ is uncountable.

I chose the following two functions:

$$f:L_a \rightarrow ℝ:x \rightarrow x$$ is injective

$$g:ℝ \rightarrow L_a:x \rightarrow a+e^x$$ is injective

It follows that a bijection exists between both sets, so $$L_a$$ is uncountable.

EDIT: corrected use of SBC, it should now work.

• You are using Schröder-Bernstein-Cantor Theorem wrongly. Jul 6 at 7:26
• Doesn't Schroder-Bernstein require injections in both directions. Jul 6 at 7:27
• @Cpc you're right. Thanks for pointing that out, I noted it wrong in my notes. Jul 6 at 7:30
• Yeah, your two functions give the same inequality. You don't need a bijection. Just the other direction. Jul 6 at 7:31
• Do you know that an interval is uncountable? Jul 6 at 7:36

The function $$g$$ doesn't make sense. For instance, $$g(-1)\notin L_a$$.

You can simply say that the function$$\begin{array}{ccc}\Bbb R&\longrightarrow&L_a\\x&\mapsto&a+e^x\end{array}$$is a bijection. Therefore, $$L_a$$ is uncountable.

• Thanks! I wanted to solve this using Schröder-Bernstein-Cantor, but I think I have it corrected now. Jul 6 at 7:35

Let $$R_a:= \{x \in \mathbb R: x and define $$f: L_a \to R_a$$ by $$f(x)=2a-x.$$

Then $$f$$ is a bijection. Suppose that $$L_a$$ is countable, then $$R_a$$ is countable, hence

$$\mathbb R= L_a \cup R_a \cup \{a\}$$

$$\tan:(-\pi/2,\pi/2)\to\Bbb R$$ is a bijection.
But any two (finite) open intervals are isomorphic (there's a bijection between them): $$(c,d)\leftrightarrow(a,b)$$ by $$x\to \dfrac{(d-x)}{(d-c)}a+\dfrac{(c-x)}{(c-d)}b$$.
Thus $$L_a$$ is uncountable, because it contains an interval.
• Your bijection doesn't work when $b=\infty$ Aug 3 at 13:11