5

I have the following topology : $$\tau= \Bigl\{U\subseteq \mathbb{R}^2: (\forall(a,b) \in U) (\exists \epsilon >0) \bigl([a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq U\bigr)\Bigr\}$$

Are these a basis for the previous topology:

$\beta_1= \{[a,a+\epsilon] \times [b-\epsilon, b+\epsilon]\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}$

$\beta_2= \{[a,a+\epsilon) \times [b-\epsilon, b+\epsilon)\subseteq \Bbb R^2: (a,b)\in \Bbb R^2, \epsilon>0 \}$

The first one is obviously a basis for $\tau$ because of the definition of $\tau$ and I would say that the second is also a basis, because $[a,a+\epsilon) \times [b-\epsilon, b+\epsilon) \subseteq [a,a+\epsilon] \times [b-\epsilon, b+\epsilon] $

Is it correct? what do you think?

Blanca
  • 741

2 Answers2

2

Hint. What definition of “basis” are you using? There are a few variations of the definition around, but all the ones I know start by requiring that the sets in the basis should themselves be open sets of the topology. It’s easy to jump straight to checking the other properties, but it’s important not to forget that first condition.

Peter LeFanu Lumsdaine
  • 9,434
2

Note that no element in $\beta_1$ is actually an open set in the topology $\tau$. Consider the element $A$ in $\beta_1$ with $a=b=0$ and $\epsilon=1$. The point $(1,1)\in A$ but there exists no $\epsilon>0$ such that $[1,1+\epsilon]\times[1-\epsilon,1+\epsilon]\subset A$ because $(1,1)$ is on the boundary of $A$.

For $\beta_2$, consider the same $a,b,\epsilon$ but look at the point $(0,0)$.

Dan Rust
  • 30,108
  • So none of them are a basis? I don't understand why the first one isn't. I have to prove that for any $U\in \tau$ any $(a.b)\in U$, exists $B_{(a,b)}\in \beta_1: (a,b)\in B_{(a,b)}\subseteq U$ And why does not satisfy this? – Blanca Oct 22 '13 at 20:37
  • 1
    The elements of $\beta_1$ are not open in $\tau$, which is a necessary condition for $\beta_1$ to be a basis of $\tau$ (by any normal definition of basis anyway). Check my example $A$ above. It does not satisfy the conditions necessary to be an element of $\tau$. – Dan Rust Oct 22 '13 at 20:45
  • 1
    The definition you mention is not the usual definition of a basis of a topology. It's not even equivalent to the normal definition because otherwise the set of closed intervals would form a basis for the usual topology on $\mathbb{R}$. – Dan Rust Oct 22 '13 at 20:57
  • Which is the problem with the point (0,0)?? Taking a=b= 0 and $\epsilon=0$ we get that A=[0,1)x[-$\epsilon, \epsilon$], and for the point (0,0), [0,$\epsilon$]x [-$\epsilon$, $\epsilon$] is part of A isn't it? – Blanca Oct 22 '13 at 23:01
  • 1
    @Blanca: You should look at the point $(0,-1)\in[0,1)\times[-1,1)$. – Stefan Hamcke Oct 22 '13 at 23:26