Notes - Metric Spaces MT23, Open and closed sets
Flashcards
What does it mean for a subset $U \subseteq X$ to be open in $X$?
Under what metric is every set open?
The discrete metric.
What does it mean for a subset $F \subseteq X$ to be closed in $X$?
$F^C$ is an open subset of $X$.
Give examples of sets that are:
- Not open and not closed
- Not open and closed
- Open and not closed
- Open and closed
- Not open and not closed: $(0, 1]$
- Not open and closed: $[0, 1]$
- Open and not closed: $(0, 1)$
- Open and closed: $\mathbb R$
What result lets you form new open sets from the union of existing open sets?
For any indexing set $I$,
\[\bigcup_{i \in I} U_i\]is an open set.
What result lets you form new open sets from the intersection of existing open sets?
For any finite indexing set $I$,
\[\bigcap_{i \in I} U_i\]is a closed set.
Can you give an example of an infinite intersection of open sets such that the result is not open?
has intesection $\{0\}$ which is not open in $\mathbb R$.
What result lets you form new closed sets from the intersection of existing closed sets?
For any indexing set $I$,
\[\bigcap_{i \in I} U_i\]is a closed set.
What result lets you form new closed sets from the union of existing closed sets?
For any finite indexing set $I$,
\[\bigcup_{i \in I} U_i\]is a closed set.
What is true about the pre-image of an open set under a continuous function?
It is also open.
What is true about the pre-image of a closed set under a continuous function?
It is also closed.
Suppose we have $\mathcal U \subseteq Y \subseteq X$ where $X$ is a metric space, and $\mathcal U$ is an open in $Y$. What condition allows you to link to the open/closed subsets of $X$?
and similarly for closed.
Quickly justify why for any indexing set $I$,
\[\bigcup_{i \in I} U_i\]
is an open set and for any finite indexing set $I$,
\[\bigcap_{i \in I} U_i\]
is a closed set.
For the union, if $x \in U _ i$ for some $i$, then you can just take the $\varepsilon$ that works for $U _ i$.
For the intersection, if $x \in U _ i$ for some $i$, then you can take the minimum $\varepsilon$ over all $U _ i$.
Quickly prove that every open ball in a metric space is an open set.
Suppose the ball is $B(a, \varepsilon)$. Let $x \in B(a, \varepsilon)$. Then $d(x, a) < \varepsilon$, so there exists $\varepsilon’ > 0$ such that $d(x, a) < \varepsilon - \varepsilon’$. Then $B(x, \varepsilon’) \subseteq B(x, \varepsilon)$: for $z \in B(x, \varepsilon’)$, note that $d(z, x) \le \varepsilon’$ and so by the triangle inequality $d(z, a) \le d(z, x) + d(x, a) < \varepsilon’ + (\varepsilon - \varepsilon’) = \varepsilon$.
Quickly prove that every closed ball in a metric space is a closed set.
Suppose the ball in $\overline B(a, \varepsilon)$. We show that $\overline{B}(a, \varepsilon)^C$. Let $x \in \overline B(a, \varepsilon)^C$. Then $d(x, a) > \varepsilon$ and so we can pick some $\varepsilon’ > 0$ such that $d(x, a) > \varepsilon + \varepsilon’$. Then $B(x, \varepsilon’) \subseteq \overline B(a, \varepsilon)^C$: for $z \in B(x, \varepsilon’)$, note that $d(z, x) < \varepsilon’$, so by the triangle inequality, $d(a, z) \ge d(x, a) - d(z, x) > (\varepsilon + \varepsilon’) - \varepsilon’ = \varepsilon$.
Quickly prove that if $Y \subseteq X$ then $U \subseteq Y$ is an open subset of $Y$ if and only if there is an open subset $V$ of $X$ such that
\[U = Y \cap V\]
$U$ open subset in $Y$ implies $U = Y \cap V$ where $V$ open in $X$: For each each $y \in U$, $\exists \varepsilon _ y > 0$ such that $B _ Y(y, \varepsilon) \subseteq U$. Then
\[\bigcup_{y \in U} B_Y(y, \varepsilon_y) = U\]Define
\[V = \bigcup_{y \in U} B_X(y, \varepsilon)\]Then
\[\begin{aligned} Y \cap V &= Y \cap \bigcup_{y \in U} B_X(y, \varepsilon) \\\\ &= \bigcup_{y \in U} (Y \cap B_X(y, \varepsilon)) \\\\ &= \bigcup_{y \in U} B_Y(y, \varepsilon) \\\\ &= U \end{aligned}\]$U = Y \cap V$ where $V$ open in $X$ implies $U$ open subset in $Y$: Consider arbitrary $y \in U$. Then $\exists \varepsilon > 0$ such that $B _ X(y, \varepsilon) \subseteq V$. But then
\[B_Y(y, \varepsilon) = Y \cap B_X(y , \varepsilon) \subseteq Y \cap V = U\]so $U$ is an open subset in $Y$.