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Determine whether the collection of subsets below form a basis for a topology on $\mathbb{R}^2$.

All subsets of the form $T_{\epsilon}(x)=\lbrace (y_1,y_2) : |x_1+x_2-y_1-y_2| <\epsilon \rbrace$ for all $x \in \mathbb{R}^2$ and all $\epsilon>0$

By doing some algebra, we obtain $-\epsilon -(x_1+x_2) < y_1+y_2<\epsilon -(x_1+x_2)$, which means that the set is the region bounded by two straight lines with negative gradient and y-intercept $-\epsilon -(x_1+x_2)$ and $\epsilon -(x_1+x_2)$ respectively.

The answer given is that the collection of the subsets above does not form a basis for topology on $\mathbb{R}^2$. Why? I thought every point is contained in one of the subsets above and also intersection between any two subsets is again the region bounded by two straight lines. I don't see why it fails to be a basis.

Idonknow
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3 Answers3

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You have a sign wrong: $T_\epsilon(x)$ is the set of $y=\langle y_1,y_2\rangle$ satisfying

$$(x_1+x_2)-\epsilon<y_1+y_2<(x_1+x_2)+\epsilon\;.$$

However, you’re right: $\{T_\epsilon(x):x\in\Bbb R^2\text{ and }\epsilon>0\}$ is a base for a topology $\tau$ on $\Bbb R^2$.

In fact, let $f:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x+y$; then the open sets in $\langle\Bbb R^2,\tau\rangle$ are precisely the sets $f^{-1}[U]$ such that $U$ is an open set in the usual topology on $\Bbb R$. In fact,

$$T_\epsilon(x)=f^{-1}[(f(x)-\epsilon,f(x)+\epsilon)]\;.$$

Brian M. Scott
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I agree that it is a basis for a topology on $\mathbb{R}^2$. It is not the standard topology. I think this should be homeomorphic to the quotient of $\mathbb{R}^2$ (with the standard topology) by the relation $x \sim y$ iff $x_1 + x_2 = y_1 + y_2$.

Eric Auld
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The basis sets are $B_{\alpha,\epsilon}=\{x | |x_1+x_2 -\alpha| < \epsilon \}$, with $\alpha \in \mathbb{R}$ and $\epsilon>0$.

Let $\phi(x) = (x_1+x_2,x_2-x_1)$. We see that $\phi$ is bijective.

Note that $\phi(B_{\alpha,\epsilon}) = \{ x | |x_1-\alpha| < \epsilon \} = (\alpha-\epsilon, \alpha+\epsilon) \times \mathbb{R}$.

It is easy to see that these sets form a basis for a topology, and since $\phi$ is a bijection, the sets $B_{\alpha,\epsilon}$ form a basis for a topology.

copper.hat
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  • Mind to explain what is sp$\lbrace(1,-1) \rbrace$? – Idonknow Nov 22 '13 at 17:26
  • @Idonknow: Probably the vector space span of the set ${\langle 1,-1\rangle}$, which is the set of all $\langle x,-x\rangle$ for $x\in\Bbb R$. But it’s not that, because that only collapses the line $y=-x$ to a point. – Brian M. Scott Nov 22 '13 at 17:29
  • @Idonknow: My earlier answer was wrong & sloppy. I should have written something like: Let $V= {(x,-x)}{x \in \mathbb{R}}$, and $X = \mathbb{R}^2 / V$, the quotient space. Then the open sets in your question are precisely the sets $\cup{L \in U} L$, where $U$ is open in $X$. I changed my answer to avoid any mention of quotient spaces. – copper.hat Nov 22 '13 at 18:10