# Sequence of topological spaces

A friend of mine did an exercise where a part of the text was:

In $\mathbb{R}^3$, with euclidian topology, we consider $X=\mathbb{S}^2 \setminus \{ N \}$, where $N= (0,0,1)$ and $E=\{(x,y,z) \in \mathbb{S}^2 \mid z=0 \}$. Let be $Y= X/E$ the quotient space obtained from the contraction of $E$ to one point and let $\pi\colon X \rightarrow Y$ be the canonical projection

Now a game started in my mind.

I know that $Y$ is homeomorphic to two spheres, one of which has a point removed. Now let $X=X_0$ and $Y=X_1$. For each sphere of $X_1$, I'll do the same thing that i did to $Y$ so I'll have $X_2$ that is homeomorphic to $4$ spheres where one of these is without a point. I know that $\mathbb{S}_2 \setminus \{ N \}$ is homeomorphic to a close disk $D^2$. So I create a sequence $\{X_i\}_{i \in \mathbb{N}}$ of topological spaces where each $X_i$ is homeomorphic to $2^i-1$ spheres tangent where the last one is tangent to a disk. The question is:

Is it possible to determinate the space $$\lim_{i \to \infty}X_i$$

Is it a topological space? Does this sequence converges?

Are we in the space of the topological space? Which kind of space is this?

Sorry for my bad english, i hope someone will fix my errors.

Thank you

• What do you mean by limit? – Dan Rust Aug 31 '15 at 21:08
• I don't know. It's possible to consider the limit of this sequence? Or is a my "invention"?.. Another question.. Where are we?.. In which space?.. What is the space of the topologycal space?.. We can define a norm?.. – Skills Aug 31 '15 at 21:14
• No, it does not make sense to take the limit of topological spaces without any additional data. It is possible to take a limit of compact metric spaces. See Gromov-Hausdorff metric. (Just as with any sequence of points in a metric space, though, this limit may not exist!) – user98602 Aug 31 '15 at 21:15
• Hmm. Unless I misunderstood we have compatible maps $\pi_{ij}:X_i\to X_j$ whenever $i\le j$, so a direct limit is probably what you want. – Jyrki Lahtonen Aug 31 '15 at 21:17
• I don't have anything to add to that Wikipedia page. Do observe that the end result will have the final topology. My guess is that you will be contracting all the latitudes that have a $z$-coordinate of the form $k/2^n$ for some integers $k,n$ to single points. I do not have a clear intuitive picture of what the topology will be like. Looks very weird :-) – Jyrki Lahtonen Aug 31 '15 at 21:23

You could so something very similar to you example, except start with a copy of the circle in the plane, glue together the east and westmost points to get $X_1$, then form $X_2$ which is four circles, and so on and so forth.
In the same way that you can think of the line with two origins as being two copies of the real line line glued together point for point except for the origin, I think the direct limit of these circular $X_i$ will be $$(I\times\{0\}\sqcup I\times \{1\})/{\sim}$$ where $(x,0)\sim (y,1)$ if and only if $x=y=k/2^n$ for some $n\geq 0$ and integer $0\leq k \leq 2^n$.
I think the best way I could try to imagine the direct limit in the spherical case of the $X_i$s is something like the above defined space 'rotated about its axis', so instead of every point not of the form $k/2^n$ having exactly two representatives, we now have an entire circle worth of representatives, all points of which have open neighbourhoods which will always intersect non-trivially.