Why are the topological dimensions of spatial\timelike\null infinities 0\0\2 respectively? I have a question regarding spatial, timelike and null infinities in Minkowski spacetime. I shall start with an explanation of  my intuitive understanding of these concepts, and then proceed to my actual question:
My Intuitive Understanding
Null infinity: I assume that I am currently in the (arbitrary) origin of flat spacetime. I trace the past history of a sphere, which is currently a just a point in the origin and whose radius grows at the rate of the speed of light as time goes to the past. As I look farther back in the past, this sphere becomes larger and larger. When I take time to approach minus infinity, then this is past null infinity. It is topologically a 2-sphere (since it was a sphere all along the way the process...).
Likewise, future null infinity is the limit at infinite time of a sphere expanding towards to the future at the speed of light, and it is also a 2-sphere.
Timelike infinity: Light from a specific very early time could have arrived at the origin only if it started anywhere on the appropriate sphere. In contrast, a massive particle in inertial motion could have arrived from anywhere inside the sphere (the exact point determined by the momentum). So it seems to me that past timelike infinity: should topologically be of dimension not lower than that of past null infinity - 2. Likewise, future timelike infinity: should have dimension not lower than 2.
spatial infinity By the same argument I would think that there should be a past spatial infinity and a future spatial infinity, with dimensions not lower 2.
My Question
As demonstrated in a Penrose-Carter diagram, the past and future timelike infinities  are two distinct points (0-dimensional). Spatial infinity is just one point (0-dimensional), and there is no separation into future and past spatial infinities. And the future and past null infinities are 3-dimensional null surfaces $\mathbf{R} \times S^2$ . Obviously my intuitive description above is very wrong. So What is a better way to imagine these infinities?
(I take this from: Carroll, Sean M. Spacetime and geometry. An introduction to general relativity. Vol. 1. 2004.)
 A: 
As demonstrated in a Penrose-Carter diagram, the past and future timelike infinities  are two distinct points (0-dimensional). 

Because we made them that way. Notice in page 475 the text says that Minkowski space is technically only the interior of that triangle, none of the sloping edges on the right, and none of the three corners. Some people draw it as a diamond so you don't have to include that line on the left.
Let's do an example where we add stuff to a manifold. Imagine your manifold is the inside of a disk, i.e. every point where $z=0$ and $x^2+y^2<1.$ 
Now there are at least two ways to add points to that manifold to make it be topologically compact. One way is to make it the closed disk, i.e. every point where $z=0$ and $x^2+y^2\leq 1.$ 
Another way is to make it be (topologically) a spherical shell surface. For instance you make the north pole be a point and you identify each non north pole point as a point on your original manifold.
Let's be clear that this is a topological addition. Another way to make the set topologically compact is just to add some more charts and some more points. And that's where you can see the difference between the two. The new charts for the disk case cover the open sets intersected with the disk. The new charts for the spherical shell (the one point compactification) could cover the intersections of the open sets that contain the whole circle $x^2+y^2=1.$ And then you have to think of the whole circle as one point. Or you can imagine mapping each point in $x^2+y^2<1$ onto the surface of a sphere of circumference 2 and having the origin be the south pole and each radial line be a line of longitude. Then add the one new point, the north pole, and the new charts will cover the intersections of the open sets with the spherical shell surface. So the point is that any chart containing the north pole has some distance $d$ where the chart in question contains every point in the original disk $z=0,$ and $x^2+y^2<1$ that is $d$ or more away from the origin, i.e. an open set that contains all the points in the origonal disk that were inside some open set where the open set contained the entire circle.
In summary we added points to Minkowski space. We could have added different points, different numbers of points, different topologies of points, and added them differently in general. And then we drew the new spacetime, with the extra points in the new picture.
The point is that we had multiple ways to add some points and add new charts to the manifold that didn't change anything about the existing points. We can define the distance functions, charts, metric, and so on with no changes. And now you have a manifold with boundary.
And to be clear, these two new points, the timelike infinities are single points because we made them that way. Just like with the disk we could have had more or less points. We could have made them the same point if we wanted. That would be like taping figure H.4 onto a cylinder so that $i^+$ and $i^-$ touch.

Spatial infinity is just one point (0-dimensional), and there is no separation into future and past spatial infinities.

A radial geodesic of $t=const$ in Minkowski space heads right towards spatial infinity. And we made it be one point, again by our choice. This is exactly analogous to adding just that one point to the disk rather than the whole edge. In page 476 the text makes it sound more inevitable than it is, the curves that go from the origin to spatial infinity have an infinite length, so there was plenty of room for different compactifications. By making all those curves go to the same point in the compact i fixation your textbook chooses you can just talk about a curve going to infinity B without worrying about which direction or which time.

And the future and past null infinities are 3-dimensional null surfaces $\mathbf{R} \times S^2$ . 

Yes, and again we choose to make them that way we kept every radial null geodesic separate as we added more points to our space time diagram. That's what figure H.2 is all about.

What is a better way to imagine these infinities?

Its better to imagine your spacetime is sitting inside a larger spacetime. It takes an infinite amount of clock ticks or rulers to get out of the part you are interested in (the original Minkowski space). But since there are multiple ways to put the same Minkowski spacetime into different larger spacetime, you can't see how that works unless you visualize the larger space.
So look at Figure H.3, but first we are going visualize $\mathbb S^3$ the hard way (but easier for me to describe with text). Now $R$ is like a new angle on a unit 3 sphere, so when it is zero or $\pi$ you are at a north or south pole of the unit 3 sphere and when it is in between it is like being at a latitude you have a little sphere of smaller than unit radius (until you get to $R=\pi/2$ where it is of unit radius). So you have $R$ which is like latitude and then for each $R=const$ (except $R=0$ or $R=\pi$) you get a whole sphere of radius $\sin R$.
Now look at that cylinder, I think they drew too much. You should only have half the cylinder, say the right half. Slice it horizontally so you see a circle. Now imagine that circle as like a clock face and we want it from nine o'clock to three o'clock. That's the half we want. If you look at the cylinder in the picture there is a vertical line labeled $R=0$ and another one labelled $R=\pi$ that is 3 o'clock and 9 o'clock. We want that right half. So now you can think of $R$ as measuring that path length from $R=0$ to $R=\pi$ as you go around on the right. Now for each $R$ on the right hand side you can drop a line from that point to the plane continuing the line $R=0$ and $R=\pi.$ This tells you the size of the 2 sphere at that $R.$
This seems way more complicated than it needs to be. If you move down from the north pole you get larger and larger circles as your lines of constant latitude. Here as you increase $R$ you get larger and larger 2 spheres. 
So go back to the clock. At 3 o'clock ($R=0$) you have just one point (the north pole). As you move from 3 o'clock ($R=0$) to 9 o'clock ($R=\pi$) you have a whole sphere (not ball, just the spherical shell) of radius just large enough that it could touch the straight line on your clock from from 3 o'clock to 9 o'clock, i.e. radius $\sin R.$
So you could cover the whole upper half of your clock with vertical straight lines. These straight lines tell you how big a sphere you have at that clock position. So imagine a clock hand and the hour it points to has a sphere there of radius equal to the vertical line dropping down from that time on the clock.
So far all we've done is visualize $\mathbb S^3.$ which in the picture of the cylinder is just that circular segment of arc from $R=0$ to $R=\pi$ on the right. The text drew it as just a 1d circular arc, we had to visualize it as a whole $\mathbb S^3$ which we did by imaging a bunch of different sized ball on surfaces each getting bigger then smaller.
And a time lapse picture could be helpful too. You could imagine taking a bunch of constant latitude slices of the earth. Then placing then in a movie and you'd see a point (the north pole) then circles that get larger and larger, then reach unit size, then get smaller and smaller until they become a circle again. So the movie all fits inside a disk.
Similarly you can make a movie of $\mathbb S^3.$ It all takes place inside a unit ball. At first you have the point at the center $R=0$ (but think of $R$ as like lattitude, not as radius). Then later you get a spherical shell and it gets larger and larger until it has a surface area of $4\pi$ square units. Then it starts to shrink. These shrinking regions are different than the previous ones. And eventually they come to a point, the point $R=\pi$ (remember it is an angle like latitude). The way you move around to nearby points in $\mathbb S^3$ is to move to nearby film times and nearby points. So $R=0$ and $R=\pi$ are about as far away as you can be.
So we have $\mathbb S^3$  (which is also just the unit vectors in $\mathbb R^4$ if you already can visualize 4d easily). So now to make the "cylinder" we add time. This is easy, take figure H.4 make sure the left line is twice as long as the triangle is wide and then tape it to that cylinder with the line on the line $R=0.$
What you have here is just a line segment of $T$ that starts out length $2\pi$ and gets proportionally smaller as you go around the right part of the cylinder.
Now your clock ticks an infinite number of times getting to the edge of that triangle and you need an infinite number of rulers to get around the cylinder so all these new points on the edge of the triangle weren't in the original spacetime except $R=0$ which for $T$ with $|T|<\pi$ is just the origin of Minkowski space.
However all the issues about some things being a point versus a sphere were about how $R$ is an angle when we embedded space into $\mathbb S^3$ and we did it so that $R$ was an angle.
A good example for that is stereographic projection, put a flat surface on the north pole and make that the origin. The points on the flat surface that are far from the origin get mapped near the south pole when you map by where the point hits the sphere on its way to the south pole. All the points far from the origin are getting mapped to near the south pole of $\mathbb S^3$ (the unit vectors in $\mathbb R^4$). But the number of rulers you need didn't change. We are making just one point because we felt like it. It doesn't relate to spacetime or time, its just way we spatially compactified.
And yes this is all unphysical since I spent all the time talking about the larger imaginary spacetime of which Minkowski space is just a part. But that is how you get the infinities.
And that other half of the cylinder is useful if you want the diamond instead of a triangle and have negative $R$ which could be just continuing the radial line through the origin so it isn't a new spatial region it just makes $R=0$ less weird but $R=0$ is still a coordinate singularity.
