Proving that a non-empty countably infinite subset of $\mathbf{R}$ is not open.

Question

I am require to prove the following propostion.

Proposition. Given a non-empty countably infinite subset of $\mathbf{R}$ say $F$, it must be the case that $F$ is not open.

despite thinking on how to proceed here i am a little stuck. Could you please provide some hints to get me going.

Please refrain from providing the complete proof. I am merely looking for some ideas to get me started.

Answer 1

Actually, we can show this even without the countably infinite requirement. We can show the stronger statement:

Every non-empty open subset of $\mathbb R$ is uncountably infinite.

This is quite easy to prove. Let $x\in A\subseteq\mathbb R$. Hence, you need to show that there must be an interval of points around $x$ that is a subset of $A$ in order for $A$ to be open. You can show that any interval of points in $\mathbb R$ is uncountably infinite. This would constitute a proof.

Answer 2

1) Suppose $F$ is not empty and open. 2) Take an element $a\in F$. 3) Use the definition of open set for the point $a$. 4) You should now be looking at a particular subset of $F$. 5) Recognise that that subset has lots and lots of elements.

Answer 3

A nonempty open subset of $\mathbb{R}$ must contain, as a subset, an interval $(a,b)$, where $a,b\in\mathbb{R}$, and $a < b$.

Thus, it suffices to show that $(a,b)$ is uncountable.

But you know that the interval $(0,1)$ is uncountable.

Thus, if you can find a bijection from $(a,b)$ to $(0,1)$, you're done.