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Find a subbasis for metric topology of $\mathbb R^2$ such that it's not a basis itself .

I know that metric topology is the collection of open balls in $\mathbb R^2$ I also know how to use the interval in $\mathbb R$ to find basis and subbasis but I'm not sure how I can do that in $\mathbb R^2$.

In $\mathbb R$, I just simply take $U=\{(a,a+1): a\in \mathbb R\}$ this is not basis because $(\frac{1}{2},\frac{3}{2}) \in \mathbb R$ but $(0,1) \cap (\frac{1}{2},\frac{3}{2})= (\frac{1}{2},1) \not\in U$. But in $\mathbb R^2$ if I say $U=\{((x,y),(x+1,y+1)): x,y\in \mathbb R\}$, it just doesn't make any sense.

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So in $\Bbb R$, you took open intervals of length $1$. In $\Bbb R^2$, you could perhaps take the interiors of squares of... – David Mitra Mar 13 '14 at 12:10
center $(0,0)$ and the length of sides is 2? I tried that, but I still can't see why it's subbasis but not bassis. – Diane Vanderwaif Mar 13 '14 at 12:22
Take the collection of the interiors of all squares of area $1$. So the collection of all sets of the form $(x,x+1)\times (y,y+1)$. (It seems this is what you were trying to do.) – David Mitra Mar 13 '14 at 12:23
ok, I see that the intersection of 2 squares of area 1 is not a square of area 1 so this is not a basis, but I still don't understand that why it's a subbasis? – Diane Vanderwaif Mar 13 '14 at 12:30
Each subbasis element is open in $\Bbb R^2$. The intersections also contain all "open squares" of area at most $1$. So, the basis generated gives the usual topology of $\Bbb R^2$. – David Mitra Mar 13 '14 at 12:35
up vote 0 down vote accepted

All sets of the form $(a,b) \times \mathbb{R}$ or $\mathbb{R} \times (a,b)$, where $(a,b)$ is a usual interval in $\mathbb{R}$ will do. This collection doesn't form a base, as all members are unbounded in one direction, so they don't fit into normal open balls. But their intersections are the open squares, which do form a base, as is well-known.

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