Notes - Metric Spaces MT23, Interiors and closures
Flashcards
What is the interior $\text{int}(S)$ of a subset $S \subseteq X$ of a metric space?
The union of all open subsets of $X$ contained in $S$.
What is the closure $\bar S$ of a subset $S \subseteq X$ of a metric space?
The intersection of all closed subsets of $X$ containing $S$.
What is the boundary $\partial S$ of a subset $S \subseteq X$ of a metric space?
What does it mean for a subset $S \subseteq X$ of a metric space $X$ to be dense?
What is $\text{int}(\mathbb Q)$?
Can you give the charactersiation of the closure of $S \subseteq X$ in terms of open balls?
Suppose $S \subseteq X$ is a subset of some metric space $X$. Then $a \in \bar{S}$ if and only if every open ball $B(a, \varepsilon)$ contains a point of $S$.
Can you give the characterisation of the closure of $S \subseteq X$ in terms of limits?
Suppose $S \subseteq X$ is a subset of some metric space $X$. Then $a \in \bar{S}$ if and only if there exists a sequence of elements of $S$ with $\lim _ {n \to \infty} x _ n = a$.
Can you give the characterisation of $S \subseteq X$ being a closed set in terms of limits?
The limit of every convergent sequence of the elements of $S$ lies in $S$.
Can you give an example of when $\bar{B}(a, \varepsilon) \ne \overline{B(a, \varepsilon)}$?
Take any two element set and the discrete metric.
Can you give the characterisation of $y _ 0 \in \overline Y$ in terms of balls?
Can you give the characterisation of $y _ 0 \in \text{int} Y$ in terms of balls?
Can you give the characterisation of $y _ 0 \in \partial Y$ in terms of balls?
Can you give the characterisation of $\overline Y$ in terms of the set of limit points $L(Y)$?
Can you give the characterisation of $\overline Y$ in terms of the boundary $\partial Y$?
Quickly prove that
\[\overline A \cup \overline B = \overline{A \cup B}\]
where $\overline A$ denotes the closure of $A$.
- $\overline A \cup \overline B \subseteq \overline{A \cup B}$: Because $A \subseteq A \cup B$ and $B \subseteq A \cup B$, and closure preserves inclusions, $\overline A \subseteq \overline{A \cup B}$ and $\overline B \subseteq \overline{A \cup B}$, so $\overline A \cup \overline B \subseteq \overline{A \cup B}$.
- $\overline{A \cup B} \subseteq \overline A \cup \overline B$: We have $A \cup B \subseteq \overline A \cup \overline B$, and then taking the closure of both sides shows $\overline{A \cup B} = \overline A \cup \overline B$, where the RHS is unaffected since it is already closed.