Notes - Analysis II HT23, Limit points
Flashcards
Can you give the definition of a limit point in full?
$p \in \mathbb{R}$ is a limit point of $E \subseteq \mathbb{R}$ if $\forall \varepsilon > 0 \text{ } \exists x \in E \text{ s.t. } 0 < \vert x-p \vert < \varepsilon$.
What does it mean for a set $E \subseteq \mathbb{R}$ to be closed?
It contains all its limit points.
If $E \subseteq \mathbb{R}$ is closed, then what is true about $E^C$?
It is open.
What does it mean for a set $E \subseteq \mathbb{R}$ to be open?
Can you give the characterisation of limit points using sequences?
Quickly justify the sequences characterisation of limit points, i.e.
\[\forall \varepsilon > 0 \text{ } \exists x \in E \text{ s.t. } 0 < |x-p| < \varepsilon \iff \exists (p_n),\text { } p_n \ne p, \text{ } p_n \to p, \text{ } p_n \in E\]
$\implies$: Take $\varepsilon = \frac 1 n$, then $\exists x _ n, x _ n \to p, x _ n \in E$. $\impliedby$: For all $\varepsilon > 0$, then $\exists p _ N$ such that $ \vert p _ N - p \vert < \varepsilon$, and also they are not equal.
Proofs
Prove the equivalance of the following two definitions for $p$ being a limit point of a set $E \subseteq \mathbb{R}$:
$\forall \varepsilon > 0 \exists x \in E \text{ s.t. } 0 < \vert x-p \vert < \varepsilon$
and
$\exists (p _ n), p _ n \in E, p _ n \ne p, p _ n \to p$
$\forall \varepsilon > 0 \exists x \in E \text{ s.t. } 0 < \vert x-p \vert < \varepsilon$
$\exists (p _ n), p _ n \in E, p _ n \ne p, p _ n \to p$
Todo?