[위상수학] Munkres 12~13 Exercise 풀이

 

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1. Let $X$ be a topological space; let $A$ be a subset of $X$. Suppose that for each $x \in A$ there is an open set $U$ containing $x$ such that $U \subset A$. Show that $A$ is open in $X$.

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Sol) Let $\mathscr{T}$ is a topology of given set $X$. What we want to show is $A \in \mathscr{T}$.

Let $I$ is an index set for open sets $U$ and $\alpha \in I$.

 

By definition of topology, $\bigcup_{\alpha \in I} U_{\alpha}$ is also an open set. Using this fact, we want to show that $\bigcup_{\alpha \in I} U_{\alpha} = A$.

 

①$\bigcup_{\alpha \in I} U_{\alpha} \subset A$

Take arbitrary $x \in \bigcup_{\alpha \in I} U_{\alpha}$. Then, we know that there exists at least one $U$ such that $x \in U$. Since $U \subset A$, $x \in A$.

 

②$A \subset \bigcup_{\alpha \in I} U_{\alpha}$

Take arbitrary $x \in A$. Then, we assumed that for that $x$, there exists $U$ such that $x \in U$. Since $U \subset \bigcup_{\alpha \in I} U_{\alpha}$, $x \in \bigcup_{\alpha \in I} U_{\alpha}$. $\blacksquare$

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2. Consider the nine topologies on the set $X = \{ a, b, c \}$ indicated in Example 1 of $\S$12. Compare them; that is, for each pair of topologies, determine whether they are comparable, and if so, which is the finer.

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Sol) Let $\mathscr{T_1} = \{ \varnothing , X \}$, $\mathscr{T_2} = \{ \varnothing , \{ b\} , X \}$, $\mathscr{T_3} = \{ \varnothing , \{ a, b\} , X \}$, $\mathscr{T_4} = \{ \varnothing , \{ a \} , \{ a, b\}, X \}$, $\mathscr{T_5} = \{ \varnothing , \{ a \} , \{ b, c\}, X \}$, $\mathscr{T_6} = \{ \varnothing , \{ a \} , \{ b\} , \{ a, b\} , X \}$, $\mathscr{T_7} = \{ \varnothing , \{ b \} ,\{ a, b\} ,\{ b, c\} , X \}$, $\mathscr{T_8} = \{ \varnothing , \{ b \} , \{ c \} ,\{ a, b\} ,\{ b, c\} , X \}$, $\mathscr{T_9} = \{ \varnothing , \{ a \} , \{ b \} , \{ c \} , \{ a, b\} , \{ b, c\} , \{ c, a\} ,X \}$.

 

Since $\mathscr{T_1}$ is trivial topology, all pairs of $\mathscr{T_i} (i = 2, \cdots , 9)$ with $\mathscr{T_1}$ are comparable, and $\mathscr{T_1} \subset \mathscr{T_i}$.

 

Similiarly, $\mathscr{T_9}$ is discrete topology, $\mathscr{T_i} \subset \mathscr{T_9} ( i = 1, \cdots, 8)$.

 

So, consider the cases of pair except with $\mathscr{T_1}$ and $\mathscr{T_9}$. 

 

$\mathscr{T_2} \subset \mathscr{T_6}, \mathscr{T_7}, \mathscr{T_8}$ and the other pairs with $\mathscr{T_2}$ are not comparable.

 

$\mathscr{T_3} \subset \mathscr{T_4}, \mathscr{T_6}, \mathscr{T_7}, \mathscr{T_8}$ and the other pairs with $\mathscr{T_3}$ are not comparable.

 

$\mathscr{T_3} \subset \mathscr{T_4} \subset \mathscr{T_6}$ and the other pairs with $\mathscr{T_4}$ are not comparable.

 

$\mathscr{T_5}$ are not comparable with any $\mathscr{T_i}$s.

 

$\mathscr{T_2}, \mathscr{T_3}, \mathscr{T_4} \subset \mathscr{T_6}$ and the other pairs with $\mathscr{T_6}$ are not comparable.

 

$\mathscr{T_2}, \mathscr{T_3}\ \subset \mathscr{T_7} \subset \mathscr{T_8}$ and the other pairs with $\mathscr{T_7}$ are not comparable.

 

$\mathscr{T_2}, \mathscr{T_3}, \mathscr{T_7} \subset \mathscr{T_8}$ and the other pairs with $\mathscr{T_8}$ are not comparable.

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3. Show that the collection $\mathscr{T_c}$ given in Example 4 of $\S$12 is a topology on the set $X$.

Is the collection

$$\mathscr{T_{\infty}} = \{ U \mid X - U \ \mathrm{is \ infinite \ or \ empty \ or \ all \ of \ }X \}$$

a topology on $X$?

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Sol)

Recall: $\mathscr{T_c}$ is the collection of all subsets $U$ of $X$ such that $X - U$ either is countable or is all of $X$.

 

WTS: $\mathscr{T_c}$ satiesfy 3 conditions of topology.

 

①WTS : $\varnothing \in \mathscr{T_c}, X \in \mathscr{T_c}$

Take $U = X$. then $X - X = \varnothing$ is finite, so, $X \in \mathscr{T_c}$.

Take $U = \varnothing$. then $X - \varnothing = X$ is $X$ itself, so, $\varnothing \in \mathscr{T_c}$.

 

Let $\{ U_{\alpha} \}$ is an indexed family of nonempty element of $\mathscr{T_c}$.

②WTS : $\bigcup U_{\alpha} \in \mathscr{T_c}$.

We know $X - \bigcup U_{\alpha}$ $=$ $X \cap \left ( \bigcup U_{\alpha} \right )^{c}$ $=$ $X \cap \left ( \bigcap U_{\alpha}^c \right )$ $=$ $\bigcap \left ( X \cap U_{\alpha}^{c} \right )$ $=$ $\bigcap (X - U_{\alpha})$.

And we know that $X - U_{\alpha}$ is either countable or $X$.

Hence, it is intersection of countable sets or $X$, $X - \cap U_{\alpha}$ is also either countable or $X$.

Therefore, $\cup U_{\alpha} \in \mathscr{T_c}$.

 

③WTS : $\bigcap_{k = 1}^{n} U_{k} \in \mathscr{T_c}$.

We know $X - \bigcap_{k = 1}^{n} U_{k}$ $=$ $X \cap \left ( \bigcap_{k = 1}^{n} U_{k} \right ) ^{c}$ $=$ $X \cup \left ( \bigcup_{k = 1}^{n} U_{k}^c \right )$ $=$ $\bigcup_{k = 1}^{n} (X \cap U_{k}^c)$ $=$ $\bigcup_{k=1}^{n} (X - U_k)$.

And $X - U_k$ is either countable or $X$.

Since we know countable union of countable sets are countable, $X-\bigcap_{k=1}^{n} U_k$ is either countable or $X$.

Therefore, $\bigcap_{k = 1}^{n} U_{k} \in \mathscr{T_c}$.

 

Recall: $\mathscr{T_{\infty}}$ $=$ $\{ U \mid X - U \ \mathrm{is \ infinite \ or \ empty \ or \ all \ of \ }X \}$.

 

WTS : $\mathscr{T_{\infty}}$ does not satiesfy 3 conditions of topology.

 

①WTS : $\varnothing \in \mathscr{T_{\infty}}$, $X \in \mathscr{T_{\infty}}$

Since $X - X = \varnothing$, $X \in \mathscr{T_{\infty}}$. Also, since $X - \varnothing = X$ itself, $\varnothing \in \mathscr{T_{\infty}}$.

 

Let $\{ U_{\alpha} \}$ is an indexed family of nonempty element of $\mathscr{T_{\infty}}$,

②WTS : $\bigcup U_{\alpha} \notin \mathscr{T_{\infty}}$

We know $X - \bigcup U_{\alpha}$ $=$ $\bigcap (X-U_{\alpha})$ and $X-U_{\alpha}$ is infinite or empty or all of $X$.

But, $\bigcup (X - U_{\alpha})$ can be a finite set.

Suppose $X = \mathbb{R}$ and $X - U_{\alpha}$ $=$ $(-1/\alpha, 1/\alpha)$.

Since we know that $(-1/\alpha, 1/\alpha)$ is an infinite set, but $\bigcap_{\alpha = 1}^{\infty} (-1/\alpha, 1/\alpha)$ $=$ $\{ 0 \}$, so, $X - \bigcup U_{\alpha}$ is finite.

It means, in this situation, $\bigcup U_{\alpha} \notin \mathscr{T_{\infty}}$.

Therefore, $\mathscr{T_{\infty}}$ is not a topology on $X$.

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4. (a) if $\{ \mathscr{T_{\alpha}} \}$ is a familty of topologies on $X$, show that $\bigcap \mathscr{T_{\alpha}}$ is a topology on $X$. Is $\bigcup \mathscr{T_{\alpha}}$ a topology on $X$?

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Sol)

WTS : $\bigcap \mathscr{T_{\alpha}}$ satiesfy 3 conditions of topology.

 

①WTS : $\varnothing \in \bigcap \mathscr{T_{\alpha}}$, $X \in \bigcap \mathscr{T_{\alpha}}$.

Since all $\mathscr{T_{\alpha}}$ is topology, each $\mathscr{T_{\alpha}}$ contains both $\varnothing$ and $X$.

Hence, whatever we take a intersection of $\mathscr{T_{\alpha}}$, they contains $\varnothing$ and $X$.

 

Let $\{ U_{\gamma} \}$ is an indexed family of nonempty element of $\bigcap \mathscr{T_{\alpha}}$

②WTS: $\bigcup U_{\gamma} \in \bigcap \mathscr{T_{\alpha}}$

We know for all $\gamma$, $U_{\gamma} \in \mathscr{T_{\alpha}}$.

It means, if