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

"$$\mathbb R -\mathbb Q$$ 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 ?

• 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
– Alan
Jun 9 '21 at 5:29
• @Sani. does my answer answer your question ? Jun 9 '21 at 14:17

## 1 Answer

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.