# 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.

Thanks for your feedback.

• You are using Schröder-Bernstein-Cantor Theorem wrongly. Jul 6, 2022 at 7:26
• Doesn't Schroder-Bernstein require injections in both directions. Jul 6, 2022 at 7:27
• @Cpc you're right. Thanks for pointing that out, I noted it wrong in my notes. Jul 6, 2022 at 7:30
• Yeah, your two functions give the same inequality. You don't need a bijection. Just the other direction. Jul 6, 2022 at 7:31
• Do you know that an interval is uncountable? Jul 6, 2022 at 7:36

## 3 Answers

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, 2022 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

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

is countable, a contradiction.

An interval is uncountable:

$$\tan:(-\pi/2,\pi/2)\to\Bbb R$$ 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, 2022 at 13:11
• The only flaw in the reasoning here (and yes, I checked the previous edits) was the minor error at one point (since corrected) of saying the offered bijection between open intervals applied to all open intervals instead of just finite open intervals. But since the demonstration that one finite open interval was in bijection with $\Bbb R$ and thus uncountable, and the other bijection shows even small open intervals are likewise uncountable, it does follow that any set, including infinite invervals, that contains a finite open interval is also uncountable. I don't see a cause for downvoting. Oct 13, 2022 at 19:25
• @PaulSinclair thanks for weighing in Oct 13, 2022 at 22:18