- #1
Paparazzi
- 9
- 0
When considered as a subset of [itex]\mathbb{R}^2[/itex], [itex]\mathbb{Z}[/itex] is a closed set.
Proof.
We will show, by definition, that [itex]\mathbb{Z} \subset \mathbb{R}^2[/itex] is closed.
That is, we need to show that, if [itex]n[/itex] is a limit point of [itex]\mathbb{Z}[/itex], then [itex]n \in \mathbb{Z}[/itex].
I think this becomes vacuously true, since our hypothesis is false, i.e. because [itex]\mathbb{Z}[/itex] has no limit points. Is this true, or am I just being silly?
Thank you!
Edit: (I know this can be proved, again by definition, by showing that [itex]\mathbb{R}^2 - \mathbb{Z}[/itex] is open.)
Proof.
We will show, by definition, that [itex]\mathbb{Z} \subset \mathbb{R}^2[/itex] is closed.
That is, we need to show that, if [itex]n[/itex] is a limit point of [itex]\mathbb{Z}[/itex], then [itex]n \in \mathbb{Z}[/itex].
I think this becomes vacuously true, since our hypothesis is false, i.e. because [itex]\mathbb{Z}[/itex] has no limit points. Is this true, or am I just being silly?
Thank you!
Edit: (I know this can be proved, again by definition, by showing that [itex]\mathbb{R}^2 - \mathbb{Z}[/itex] is open.)