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These are very similar to my last two questions. I provide them with my thoughts so far:

$(1)$ Let $S$ be the collection of all straight lines in the plane which are parallel to the x-axis. If $S$ is a subbasis for a topology $T$ on $\Bbb R^2$ describe the open sets in $(\Bbb R^2, T)$.

Since parallel lines can't intersect each other, the elements of the basis are all lines parallel to the x-axis. This means that the open sets in the topology are all collections of lines parallel to the x-axis including contiguous blocks that look like bars parallel to the x-axis. Correct?

$(2)$ Let $S$ be the collection of all circles in the plane. If $S$ is a subbasis for a topology $T$ on $\Bbb R^2$ describe the open sets in $(\Bbb R^2, T)$.

Since we are still talking about finite intersections the elements of the basis are going to be all circles and all finite collection of distinct points that number $2x$ where $x \in \Bbb N$. Any singleton then exists in the topology because the singleton $x$ for example is just going to be $\{a,x\} \cap \{x,b\}$. So the topology is the discrete topology. Phew. Seems correct?

Sorry for the horrible writing.

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    $\begingroup$ That is correct. More directly every singleton can be looked at as the intersection of two suited circles, so is open in case (2). $\endgroup$ – drhab Mar 19 '15 at 16:00
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    $\begingroup$ @drhab So suited circles are just two circles that intersect at one point? I had never heard of the term as far as I recall. Thanks. $\endgroup$ – user224530 Mar 19 '15 at 16:37
  • $\begingroup$ If $S_1=\{(x,y)\in\mathbb{R}^{2}\mid (x-a-1)^{2}+(y-b)^{2}=1\} $ and $S_1=\{(x,y)\in\mathbb{R}^{2}\mid (x-a+1)^{2}+(y-b)^{2}=1\} $ then $S_1\cap S_2=\{ (a,b)\} $ $\endgroup$ – drhab Mar 19 '15 at 18:04
  • $\begingroup$ @drhab yeah you're right. Plugging in $\{a,b\}$ into $S_1$ and $S_2$ yields the same value. $\endgroup$ – user224530 Mar 19 '15 at 19:52

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