confusion over description for the following two topological spaces I am having trouble with the following two questions below. The difficulty I am having mainly have to do with understanding visually the described topology due to confused wordings.  
For (1), when it states: an open line segment in each direction about (x,y) that is contained in $U$. First, am I supposed to pick some patch on the x-y plane, then pick a point within the patch $U$ and consider that as an origin.  It would be like applying Euclidean transformation to the original origin.  Then through this new point of origin, I draw lines through it.  But do I do it for every direction? Also, do all the lines only supposed to stay within the patch $U$ which I have choosen?
For (2), I am confused about its description and also when trying to picture both it and the subspace topology for (a).  When it states that definitely many lines through the origin are removed.  Does it mean the following: I have an open disc with the interior coloured solid, and finite number of lines through the centre are removed. For the subspace topology in (a), suppose I have an open disc with five lines through the origin removed.  The disc intersect a line that go through the origin. Does this line from topology $D$ which intersects the open disc topology get removed also?
For (b), if I intersects the open disc topology with a circle from topology $F$, this circle which I intersects with, do I keep the centre of this circle?
1)  Let $\mathcal T=\{U \in\mathcal P(\mathbb R^2):$U = $\emptyset$ or for each $(x,y)$ $\in$ $U$ there is an open line segment in each direction about $(x,y)$ that is contained in $U$}  
a) Let $A$ denote a straight line in $\mathbb R^2$. Describe $\mathcal T_{A}$
b) Let $P$ denote a circle in $\mathbb R^2$. Describe $\mathcal T_{P}$
2) Let $\mathcal B'=\{\mathcal {B} \in\mathcal P (\mathbb R^2):$$B$ is an open disk with a finite number of straight lines through the center removed$\},$  and let $\mathcal{B}=\{B\cup\{c\}: $$B\in\mathcal{B}'$ and $c$ is the center of $B$} 
a) Let $D$ denote a straight line in $\mathbb R^2$. Describe $\mathcal T_{D}$
b) Let $F$ denote a circle in $\mathbb R^2$. Describe $\mathcal T_{F}$
Thank you in advance
 A: For part 1, the topology is the set of algebraically open sets, i.e. sets equal to their algebraic interior. Of course, every Euclidean-open set satisfies this property, but there are other sets which are not in the Euclidean topology that do not satisfy this property.
I'll construct an example. Let $C = B((1, 0); 1)$, the open disk centred at $(1, 0)$, with radius $1$. Let $D = \Bbb{R}^2 \setminus B[(2, 0); 2]$, the (open) complement of the closed disk centred at $(2, 0)$ with radius $2$. Note that the boundary of these Euclidean-open sets are circles, intersecting uniquely and tangentially at the origin. Then
$$U = C \cup D \cup \{(0, 0)\}$$
is not Euclidean open, but belongs to this topology.
I'm not going to prove that this is an example of such a set; instead I think you should draw a picture, and try to convince yourself from the geometry of the situation that this is an example.
Obviously, any point in $C \cup D$ will be in the algebraic interior, as $C \cup D$ is an open subset of $U$, and hence lies in the interior of $U$, so the only point of interest is $(0, 0)$. Indeed, this point should be in the algebriac interior, but not the interior of $U$.
Mentally consider a line, in any direction, centred at $(0, 0)$. If the line is parallel to the $y$-axis, then it lies tangent to both boundary circles, and must therefore be contained in $D \cup \{(0, 0)\}$.
If the line is not tangential, then one open ray from $(0, 0)$ must be contained in $D$, but the other open ray from $(0, 0)$ must enter into $C$ temporarily, before exiting $C$, staying in the complement of $U$ for a while, then entering in $D$ again.
The point is, while the full line is not contained in $U$, there is some non-trivial line segment in contained in $U$.
It's not hard to see that $(0, 0)$ is not in the interior of $U$. Take, for example, the circle $S[(1.5, 0); 1.5]$. This circle intersects $U$ precisely at the point $(0, 0)$, giving us points arbitrarily close to $(0, 0)$ that are contained in $\Bbb{R}^2 \setminus U$.

For part 2a, the line $D$ need not correspond to the lines removed from various balls as described in the definition of the topology. I think (again) that an example would be best to illustrate this.
Let $D = \{(x, y) \in \Bbb{R}^2 : x + y = 1\}$ and consider the open set from our topology:
$$U = \{(x, y) \in \Bbb{R}^2 : \|(x, y)\| < 1 \text{ and } y \neq x\} \cup \{(0, 0)\}.$$
That is, $U$, is the open unit disk with the line $y = x$ removed. Then $U \cup D$ is in $\mathcal{T}_D$, and is the union of two open line segments, one from $(1, 0)$ to $(1/2, 1/2)$ and one from $(1/2, 1/2)$ to $(0, 1)$.
However (hint, hint) the more interesting case is when you consider an open ball, centred on some point on the line, with the line removed! Then, the intersection is just a singleton point. This is a perfectly valid relatively open set! What does this say about the relative topology?

For part 2b, forget the centre of the circle $F$; you are only interested in how the open sets intersect $F$. In this case, try to show that relatively open subsets with respect to the topology will be relatively open with respect to the Euclidean topology too.
