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Today I began the studying of Rudin's book "Real and Complex Analysis" and have the following question:

Let $(M,d)$ be a metric space. How to turn it into topological space?

Proof: I begin the proof from the following remark.

Remark: Let $B_1, B_2, \dots, B_n$ are open balls in $(X,d)$. Suppose that $\bigcap \limits_{i=1}^{n}B_i\neq \varnothing$. Taking $x\in \bigcap \limits_{i=1}^{n}B_i$ but the last set is open hence exists open ball $U_x$ with center $x$ such that $U_x\subset \bigcap \limits_{i=1}^{n}B_i.$ Hence $\bigcup \limits_{x}U_x\subset \bigcap\limits_{i=1}^{n}B_i.$ Also if $z\in \bigcap\limits_{i=1}^{n}B_i$ then $z\in U_z$ and $\bigcap\limits_{i=1}^{n}B_i\subset \bigcup\limits_{x} U_x$. Hence $\bigcap\limits_{i=1}^{n}B_i= \bigcup\limits_{x} U_x$. Am I right?

Let $\tau_d=\{\varnothing, M, \{E\}\}$ where $E$ is the set of arbitrary unions of open balls. We have to check that $\tau_d$ is topology. Indeed,

$1)$ $\varnothing, M\in \tau_d$.

$2)$ if $E_1, \dots, E_n\in \tau_d$ then $E_1=\bigcup \limits_{i_1\in I_1}B_{r_{i_1}}(x_{i_1}), \dots, E_n=\bigcup \limits_{i_n\in I_n}B_{r_{i_n}}(x_{i_n})$ then $$E_1\cap \dots \cap E_n=\bigcup \limits_{i_1\in I_1}\dots\bigcup \limits_{i_n\in I_n}B_{r_{i_1}}(x_{i_1})\cap\dots \cap B_{r_{i_n}}(x_{i_n})$$ by above remark inner intersection is union of open balls. Hence $E_1\cap \dots \cap E_n\in \tau_d$.

$3)$ If $E_{\alpha}\in \tau_d$ for $\alpha\in I$ then $\bigcup \limits_{\alpha \in I}E_{\alpha}\in \tau_d$. This point is obvious.

I am sorry if this topic is repeated. But can anyone check my solution please?

I would be very grateful for answer\comment.

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The topology on $(M,d)$ is the set of arbitrary unions of open balls ; $\varnothing$ is the "empty union" and $M$ is the union of all of them. That is how you should prove the first axiom.

Your remark doesn't "show" that the intersection of $n$ open balls is an union of open balls (which is what you defined to mean "open"). If you assumed it because you read it in Rudin then fine, but if you wanted to prove it, you still need to work something out (hint : use the triangle inequality of the metric $d$).

The rest sounds fine. But to be technically correct, when you prove 2), you use 3) since you use the fact that an arbitrary union of open sets is open + the fact that the intersection of finitely many open balls is open (and not open balls). Try to see the distinction between the two.

Hope that helps,

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  • $\begingroup$ Thanks for answer. Is my remark false? $\endgroup$ – ZFR May 17 '16 at 22:27
  • $\begingroup$ What problem with my topology $\tau_d$? $\endgroup$ – ZFR May 17 '16 at 22:28
  • $\begingroup$ @RFZ : Your remark is not false. It's just missing a step (that you either omitted or didn't realize it was necessary; I am guessing the second since you asked me, so try thinking about my answer again). The topology $\tau_d$ is actually equal to $E$ and not to the set with three elements you wrote down (namely $\varnothing, M$ and $\{E\}$). $\endgroup$ – Patrick Da Silva May 17 '16 at 23:44
  • $\begingroup$ I reread your somment and still can't understand what step i missed. Can you show me please? $\endgroup$ – ZFR May 18 '16 at 8:11
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Property 1) is trivial, since $\emptyset$ is the union of the empty collection, while $M$ is the union of the whole collection of open balls. Property 3) is also trivial: the union of a collection of unions is a union... It is sufficient to prove Property 2) for $n=2$ (a useful simplification, at least in notation). So let $A=\bigcup_i B_i$ and $A'=\bigcup_j B'_j$; if $A\cap A'=\emptyset $ we have finished. Otherwise let $x\in A\cap A'$: then, for some $i$ and some $j$, we have $x\in B_i\cap B'_j$. The main point is that $ B_i\cap B'_j \ni x$ implies that there is a ball $B(x)$ centered in $x$ such that $B(x)\subset B_i\cap B'_j$ (prove!). Since this is true for any $ x\in A\cap A'$, you can write $$ A\cap A'=\bigcup_{x\in A\cap A'} B(x), $$ so $ A\cap A'$ is in the topology because it is union of balls.

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  • $\begingroup$ Is my topology $\tau_d$ is incorrect? It contains empty set and whole set. $\endgroup$ – ZFR May 17 '16 at 22:31
  • $\begingroup$ "I think also that there is a formal error when you say that E is the of arbitrary unions of (open) balls." What error? $\endgroup$ – ZFR May 17 '16 at 22:50
  • $\begingroup$ I rewrite the comment. Your remark (the intersection of balls is a union of balls, hence it is open) is correct. Of course the topology has to contain $\emptyset $ and $M$ (property 1)). Maybe your notation $ \tau_d=\{\varnothing, M, \{E\}\}$ is a bit redundant. I think also that there is a formal error when you say that $E$ is the set of arbitrary unions of (open) balls. If this is the definition of $E$ then, simply $\tau_d=E$. It is incorrect to write $\tau_d=\{\{E\}\}$. $\endgroup$ – guestDiego May 17 '16 at 23:04

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