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I have to compare the following topology with the usual one. Which of them is finer?

$\tau= \{U\subseteq \mathbb{R}^2:$ for any $(a,b) \in U$ exists $\epsilon >0 $ where $[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\}$

By definition, $\tau\subseteq\tau_u $ if and only if for every $U\in \tau$ implies $U\in \tau_u$

However, how can I compare them using open basis?

THANK YOU!

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  1. If $U \in \tau_u$, then for any $(a,b) \in U$, there is a basic open set $$ (a-\delta, a+\delta)\times (b-\delta', b+\delta') \subset U $$ you can take $$ \epsilon = \min\{\delta/2, \delta'/2\} $$ then $$ [a,a+\epsilon]\times [b-\epsilon,b+\epsilon] \subset U $$ Hence $U \in \tau$. So $$ \tau_u \subset \tau $$

  2. The set $$ [0,1)\times (-1,1) \in \tau\setminus \tau_u $$ Do you see why?

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  • $\begingroup$ @ Prahlad Vaidyanathar But you have to proof that for every $B\in \beta_u,$ for every $(c,d)\in B$, exists $D\in\beta : (c,d)\in D \subseteq B$ Is that epsilon correct? If I choose $(c,d)\in B=(a-\mu, a+\mu) x (b-\mu_1, b+\mu_1)$ which would be D? $\endgroup$ – Blanca Oct 22 '13 at 17:24
  • $\begingroup$ @Vaidynathar In the second point, why do you chose the set $$ [0,1)\times (-1,1) \in \tau\setminus \tau_u $$ instead of $$ [0,1]\times [-1,1] \in \tau\setminus \tau_u $$ Could you explain it please? As these sets are in $\tau$ aren't they supposed to be closed? $\endgroup$ – Blanca Oct 22 '13 at 17:35
  • $\begingroup$ @Blanca: Given $U \in \tau_u$ to show that $U \in \tau$ we just need to show that it has the defining property of the collection $\tau$, namely that for all $\langle a,b \rangle \in U$ there is an $\epsilon > 0$ such that $[ a,a+\epsilon ] \times [ b,b+\epsilon ] \subseteq U$. This is what Prahlad has done. $\endgroup$ – user642796 Oct 22 '13 at 19:33
  • $\begingroup$ @Blanca: As for your second comment, your proposed set $V = [0,1] \times [-1,1]$ does not work because it is not in $\tau$. Note that $\langle 1,1 \rangle \in V$ but there is no $\epsilon > 0$ such that $[1,1+\epsilon]\times[1,1+\epsilon] \subseteq V$. Do you see why? $\endgroup$ – user642796 Oct 22 '13 at 19:35
  • $\begingroup$ @ArthurFischer: Thank you, I understand the second.So regarding to my first comment, Prahlad has used this notation? $\tau\subseteq\tau_u $ if and only if for every $U\in \tau$ implies $U\in \tau_u$ $\endgroup$ – Blanca Oct 22 '13 at 19:51

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