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I have to prove that the following family defined in $\mathbb{R}^2$ is a topology.

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

I have started in the following way: (i) $\emptyset, X \in\tau?$ any $(a,b)\in \mathbb{R}^2 $, exists $\epsilon >0$ where $[a,a+\epsilon]\times[b-\epsilon, b+\epsilon]\subseteq \mathbb{R}^2$ so $ X=\mathbb{R}^2 \subseteq \tau$

But I don't know how to prove that the empty set is in the topology.

(ii) In the second, I have argued that taking $U_1, U_2\in \tau$, the intersection is going to be either the empty set or a set of this type: $[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]$

Is it correct?

(iii) To prove that for any $i\in I$ where $ U_i \in \tau, $ then the union of these sets is also in $\tau$ I got lost.

Could you help me please?

Thank you for your time and apologies for my poor writing.

Cameron Buie
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Blanca
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  • The empty set is in. For it is (vacuously) true that for every $(a,b)$ in $\emptyset$ (whatever) is true, since there are no such $(a,b)$. – André Nicolas Oct 22 '13 at 14:58
  • The empty set is vacuously an element of $\tau$. For the intersection of two sets in $\tau$, just take the minimum of the the two corresponding $\epsilon$ (why does this work?). For the union, taking the minimum of the set of $\epsilon$s which make sense for that element will also suffice (really you could take any of these epsilons since the union will contain all the boxes). – Dan Rust Oct 22 '13 at 15:02
  • @André Nicolas-vacuously??? I still don't know how to argue that the empty set is inside. It is absurd to write for every $(a,b)\in\emptyset$ – Blanca Oct 22 '13 at 15:27
  • Every number in the empty set is even. Every number in the empty set is odd. It is not absurd to write for every $(a,b)\in\emptyset$. – André Nicolas Oct 22 '13 at 15:32

1 Answers1

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(ii) If $(a,b) \in U_1\cap U_2$, then $(a,b) \in U_1$, so there is an $\epsilon_1$ as above, and $(a,b) \in U_2$ so there is an $\epsilon_2$ as above. Then take $$ \epsilon = \min\{\epsilon_1, \epsilon_2\} $$ Now $[a,a+\epsilon]\times [b-\epsilon,b+\epsilon] \subset U_1\cap U_2$

(iii) If $U = \cup_i U_i$ and $(a,b) \in U$, then there is an $i\in I$ such that $(a,b) \in U_i$, so there is an $\epsilon > 0$ such that $$ [a,a+\epsilon]\times [b-\epsilon,b+\epsilon] \subset U_i \subset U $$

Prahlad Vaidyanathan
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