"$\mathbb R -\mathbb Q$ is not a linear continuum"

My Attempt: I think this statement is true. Because $\mathbb R -\mathbb Q$ does not have Least Upper Bound Property.

E.g. The set $\{\mathbb R -\mathbb Q \} \cap [2 , 3]$ is bounded in $\mathbb R -\mathbb Q $. But it does not have supremum.

Can anyone please check if I have gone wrong aywhere ?

  • 2
    $\begingroup$ Looks fine to me. If you wanted to be formal, you'd need to prove why the supremum doesn't exist, but it's pretty obvious $\endgroup$
    – Alan
    Jun 9 '21 at 5:29
  • $\begingroup$ @Sani. does my answer answer your question ? $\endgroup$
    – Logic
    Jun 9 '21 at 14:17

You are correct to prove it has no irrational supremum you can show that given any irational in $a$ in $[2,3]$ there is another irrational in between $3$ and a , for example $(a+3)/2$ so the least irrational upper bound must be greater than $3$ ,let that then be $n$ but then $(n+3)/2$ is will be a lower upper bound a contradiction hence no least upper bound exists.


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