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I want to show that

Let $\mathcal{T}$ be the set of all open sets in $\mathbb{C}$. Then:

  • $\emptyset \in \mathcal{T}$ und $\mathbb{C} \in \mathcal{T}$

  • For arbitrary many $U_i \in \mathcal{T}$ is: $(\bigcup_{i \in \mathcal{I}} U_i) \in \mathcal {T}$

  • For finitely many $U_1,\ldots,U_n \in \mathcal{T}$ is: $(\bigcap_{i = 1}^n U_i) \in \mathcal {T}$


And I want to use this definition of openness in $\mathbb{C}$:

Let $U \in \mathbb{C}$.

$U$ is open $\iff \forall u \in U \,\exists r \gt 0: B_r(u) \subseteq U$

$U$ is closed $\iff U^C := \mathbb{C} \setminus U$ is open

And the open ball is defined as $B_r(u) := \{z \in \mathbb{C} \,|\, |u-z| \lt r\}$.


I get how the first point is vacuously true. But I struggle how to show the other two points. How do I make the distinction between finitely and (possibly) infinitely many open sets in the proof?

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  • $\begingroup$ How is this different in any way from the usual topology on $\mathbb{R}^2$? $\endgroup$ – Mathemagical Dec 10 '17 at 13:39
  • $\begingroup$ Mathemagical, who said it was? $\endgroup$ – fleablood Dec 10 '17 at 16:38
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For the second point: $\forall x\in \bigcup_{i\in\mathcal I}U_i$, $x\in U_i$ for some $i\in\mathcal{I}$, so there exists an open ball $B_r(x)\subset U_i$ and thus $B_r(x)\subset \bigcup_{i\in\mathcal I}U_i$. Hence $\bigcup_{i\in\mathcal I}U_i$ is open.

For the third point: $\forall x\in \bigcap_{i=1}^nU_i$, there are open balls $B_{r_i}(x)\subset U_i$. Let $r=\min_{1\leq i\leq n}r_i$, then $B_r(x)\subset \bigcap_{i=1}^nU_i$. Hence $\bigcap_{i=1}^nU_i$ is open.

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  • $\begingroup$ Thanks! Can you explain me where we make use of the fact that there are only finitely many $U_i$ in the third point. Basically I want to know why it wouldn't work with infinitely many $U_i$. $\endgroup$ – philmcole Dec 10 '17 at 16:46
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    $\begingroup$ To the proposer: In the third point: If there are infinitely many $U_i$ then there might not be a positive lower bound $r$ for the set of all $r_i .$ For example if $r_n=1/n$ then $\cap_{n\in \Bbb N}B_{r_n}(x)=\{x\}$ and no non-empty open ball of positive radius is a subset of $\{x\}$ in the usual topology of $\Bbb C.$ $\endgroup$ – DanielWainfleet Dec 10 '17 at 18:12
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Let $T$ be a topology on a set $X.$ A base (basis) for $T$ is some (any) $B\subset T$ such that each non-empty $U\in T$ is the union of a family (finite or infinite) of (one or more) members of $B.$ Of course $T$ is itself an example of a base for $T.$ So $B$ is a base for $T$ iff $T=\{\cup F: F\subset B\}.$

If $B$ is any family of subsets of $X$ then $B$ is a base for a topology on $X$ iff

(Bi). $\cup B=X.$ (Each point of $X$ belongs to at least one member of $B.$),and

(Bii). If $U,U'\in B$ and $p\in U\cap U'$ then there exists $U''\in B$ with $p\in U''\subset (U\cap U').$

(Note that (Bii) is automatically satisfied if $U\cap U'\in B$ whenever $U,U'\in B,$ as we can let $U''=U\cap U.$ For example, with the usual topology on $\Bbb R,$ the base of bounded open intervals with rational endpoints has this property.)

A metric on a set $X$ is a function $d:X\times X\to [0,\infty)$ such that

(Di). $d(x,y)=d(y,x),$ and

(Dii). $d(x,y)=0$ iff $x=y,$ and

(Diii). $d(x,y)+d(y,z)\geq d(x,z).$ (Triangle Inequality).

If $d$ is a metric on $X$ then the set of open $d$-balls is a base for a topology on $X.$ Condition (Bi) is obviously satisfied. To satisfy (Bii) we use (Diii).

The function $d(x,y)=|x-y|$ is a metric on $\Bbb C.$

Details upon request. Send $20 to .... or just leave a comment.

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