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.

  • 1
    $\begingroup$ You are using Schröder-Bernstein-Cantor Theorem wrongly. $\endgroup$ Jul 6 at 7:26
  • 1
    $\begingroup$ Doesn't Schroder-Bernstein require injections in both directions. $\endgroup$
    – Noobie
    Jul 6 at 7:27
  • $\begingroup$ @Cpc you're right. Thanks for pointing that out, I noted it wrong in my notes. $\endgroup$
    – VickyG
    Jul 6 at 7:30
  • $\begingroup$ Yeah, your two functions give the same inequality. You don't need a bijection. Just the other direction. $\endgroup$
    – Noobie
    Jul 6 at 7:31
  • $\begingroup$ Do you know that an interval is uncountable? $\endgroup$
    – Noobie
    Jul 6 at 7:36

3 Answers 3


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.

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

Let $R_a:= \{x \in \mathbb R: x<a\}$ 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.

  • 2
    $\begingroup$ Your bijection doesn't work when $b=\infty$ $\endgroup$
    – jjagmath
    Aug 3 at 13:11

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