Describing this topological space I am doing this exercise from Armstrong's Topology book, and am super confused: 

"describe each of the following spaces: [...] $\mathbb{E}^2$ with each of the circles centre the origin and of integer radius identified to a point."

First, there seems to be a grammatical error, and second, I am really confused how to approach this one, and what it is asking. I think that it is asking about circles centered at the origin and only of integer radius. (Maybe it means all circles, but only identify those which have integer radius?)
Also, when it says 'to a point' does it mean that all the circles are going to be identified with the same point?
 A: I agree that the wording is confusing.  My interpretation is that you want to consider $\mathbb{E}^2$, and using $C_n$ to refer to the circle centered at the origin with radius $n \in \mathbb{N}$, identify each $C_n$ to its own point.
Imagine holding $\mathbb{E}^2$ at the origin by your fingertips, and letting the rest of the space fall with gravity to form a cone, like this.  This way we can (visually) separate the circles of integer radius, because they are all at different heights.  Now for each circle, imagine gluing all of the points together.  The result would be something like this.
A: It seems that you're to start with the Euclidean plane. Then, every point lying on any integer-radius circle centered at the origin is supposed to be considered identical to all other such points. Put another way, let $A=\{x\in\Bbb E^2: \lvert x\rvert_2\in\Bbb Z\},$ and consider the space $\Bbb E^2$ with all points of $A$ identified.
Alternately, all points on any given circle (of integer radius with its center at the origin) are to be identified. Honestly, I can see room for either interpretation. That really is poorly phrased. Perhaps that's why the language of mathematics isn't (for example) English.
A: I think it should say:

$\mathbb{E}^2$ with each of the circles centered at the origin and of integer radius identified to a point.

This means that you put an equivalence relation on $\mathbb{E}^2$ such that each of the circles around the origin of integer radius forms a single equivalence class.  That is, you define $(x,y)\sim(z,w)$ iff either $(x,y)=(z,w)$ or there exists $n\in\mathbb{N}$ such that $x^2+y^2=z^2+w^2=n^2$.
(Alternatively, you could interpret it to mean that all of the circles together should be put into a single equivalence class, so $(x,y)\sim(z,w)$ iff $(x,y)=(z,w)$ or there exist $m,n\in\mathbb{N}$ such that $x^2+y^2=m^2$ and $z^2+w^2=m^2$.  But I think this is less likely to be the intended interpretation, because the wording with "each" suggests that each circle should separately be identified to its own point.)
A: I have a solution, from a solutions manual, of this problem that might help to clear up of your confusion:
Let $X$ be $\mathbb{E}^2$ with each of the circles centred at the origin and of integer radius identified to a point. For each $n \in \mathbb{N}$ let $C_n$
be the sphere in $\mathbb{E}^3$ of radius $1/2$ centered at $(n, 0, 0)$. 
Let $Y = \cup_{n \in \mathbb{N}}C_n$. Then $X$ is homeomorphic to $Y$.
Let $X$ be the space in question and let $f : \mathbb{E}^2 \rightarrow X$ be the identification map.
For each $n \in \mathbb{N} \:$ let $A_n = \{p \in \mathbb{E}^2 | n − 1 \leq \|p\| \leq n\}$.
For $n > 1$, $A_n$ is an annulus, so homeomorphic to a cylinder. And $A_0$ is exactly the closed unit disc.
The image of $A_0$ in $X$ is $A_0$ with its boundary circle identified to a point. We know, by the example “The identification space $B^n/S^{n−1}$” (on pages $68-69$), that $A_0$ with its boundary identified to a point is homeomorphic to $S^2$.
Likewise, we know by part $(a)$ of this problem that for $n > 0$, the image of $A_n$ in $X$ is homeomorphic to $S^2$. Now the image of $A_n$ in $X$ shares exactly one point with $A_{n−1}$ and exactly one point with $A_n$, when $n > 0$ and $A_0$ shares exactly one point with $A_1$.
Thus $X$ is homeomorphic to a countable sequence of spheres each connected to the next by one point. This is exactly the space $Y$.
