- #1
gajohnson
- 73
- 0
Hi everyone, I posted this a couple days ago and didn't get a response, so I thought I'd try again. Let me know if something about this is confusing. Thanks!
Let [itex]X[/itex] be a metric space and let [itex]x\in{X}[/itex] be any point. Prove that the set [itex]\left\{x\right\}[/itex] is closed in [itex]X[/itex]
I am defining [itex]\left\{x\right\}'[/itex] as the set of all limit points of [itex]\left\{x\right\}[/itex].
Ok, so I realize I can prove this by showing that the complement is open (simple). However, I have a question about the validity of an alternative approach:
It is clear that [itex]\left\{x\right\}[/itex] has no limit points, thus [itex]\left\{x\right\}' = [/itex]empty set. Now, the empty set is a subset of every set, thus [itex]\left\{x\right\}'\subset{\left\{x\right\}}[/itex] and so [itex]\left\{x\right\}[/itex] contains all of its limit points. Then [itex]\left\{x\right\}[/itex] is closed.
Thoughts? Does this work? If not, what am I missing? Thanks!
Homework Statement
Let [itex]X[/itex] be a metric space and let [itex]x\in{X}[/itex] be any point. Prove that the set [itex]\left\{x\right\}[/itex] is closed in [itex]X[/itex]
Homework Equations
I am defining [itex]\left\{x\right\}'[/itex] as the set of all limit points of [itex]\left\{x\right\}[/itex].
The Attempt at a Solution
Ok, so I realize I can prove this by showing that the complement is open (simple). However, I have a question about the validity of an alternative approach:
It is clear that [itex]\left\{x\right\}[/itex] has no limit points, thus [itex]\left\{x\right\}' = [/itex]empty set. Now, the empty set is a subset of every set, thus [itex]\left\{x\right\}'\subset{\left\{x\right\}}[/itex] and so [itex]\left\{x\right\}[/itex] contains all of its limit points. Then [itex]\left\{x\right\}[/itex] is closed.
Thoughts? Does this work? If not, what am I missing? Thanks!