inverse limit in the plane

Question

What stuff can I say about inverse limits regarding the mapping of $[0, 1]$ onto $[0,1]$ given by $$f(x) = \left\{ \begin{array}{ll} 2x & \mbox{if } 0 \le x \le {1\over2}\\ 1 & \mbox{if } {1\over2} \le x \le 1 \end{array} \right.$$There seems to exist a lot of things that talk about inverse limits in an algebraic setting, but I'm wondering what I can say about this example, which looks like something out of Munkres's Topology as opposed to Lang's Algebra.

Answer 1

Let $M$ be the set of all sequences $(x_1, x_2, x_3, \dots)$ of numbers from $[0,1]$ with the property that $f(x_{i+1}) = x_i$ for $i = 1, 2, 3, \dots$. If $x = (x_1, x_2, \dots)$ and $y = (y_1, y_2, \dots)$ belong to $M$, define the distance from $x$ to $y$ by $$d(x, y) = \sum_{i > 0} {{|x_i - y_i|}\over{2^i}}.$$$($Actually, $M$ is a certain subset of the Hilbert cube with the relative topology.$)$

Let $$\alpha_1 = \left\{(x_1, x_2, \dots) \in M\text{ }\Bigg|\text{ }x_1 \le {1\over2}\right\}.$$If $x \in \alpha_1$, each term of $x$ is determined by $x_1$ since $f|[0, 1/2]$ is a homeomorphism so $h: \alpha_1 \to [0, 1/2]$ given by $h(x) = x_1$ is a one-to-one transformation. It is easy to see that $h$ is continuous so $\alpha_1$ is an arc $($Arc connectedness$)$. In a similar manner, $$\alpha_n = \left\{x \in M\text{ }\Bigg|\text{ }x_n \le {1\over2}\right\}$$is an arc for each positive integer $n$ and $\alpha_1 \subset \alpha_2 \subset \alpha_3 \dots$.

Moreover, if $x$ is a point of $M$ and $x \neq (1, 1, 1, \dots)$ then $x$ belongs to $\alpha_n$ for some $n$. Let $P = (1, 1, 1, \dots)$ and $R = \alpha_1 \cup \alpha_2 \cup \alpha_3 \cup\dots$. Suppose $\epsilon > 0$. There is a positive integer $N$ such that $$\sum_{i \ge N} {1\over{2^i}} < \epsilon.$$Let $y$ be the point of $R$ such that $y_N = 1/2$. Then, $y_i = 1$ for $1 \le i < N$ and therefore $d(P, y) < \epsilon$. It follows that $P$ is in $\overline{R}$ and so $\overline{R} = M$. It is easy to see that $R$ is connected so $M$ is connected.

Further, $M$ is closed since if $x$ is a limit point of $M$ then, for each positive integer $i$, there is a point $y^i$ fo $M$ such that $d(x, y^i) < 1/2^i$. Then, if $1 \le j \le i$, $x_j = y_j^i$ and consequently, $f(x_{k+1}) = x_k$ for $1 \le k \le i$. It follows that $x \in M$.

Finally, we observe that $M$ is compact. One way to see this is to note that $M$ is a closed subset of the Hilbert cube and, so, is compact. Alternatively, a notationally horrendous argument would involve taking a sequence $x^1, x^2, x^3, \dots$ of points of $M$ and getting a subsequence whose first terms converge to a point $x_1$ of $[0, 1]$. From this subsequence get a subsequence whose second terms converge to a point $x_2$ of $[0, 1]$ and notice by the continuity of $f$ that $f(x_2) = x_1$. Continuing in this manner, for each positive integer $n$, we get a subsequence whose first $n$ terms converge to points $x_1, x_2, x_3, \dots, x_n$ and $f(x_{i+1}) = x_i$ if $1 \le i < n$. Taking the subsequence of $x^1, x^2, \dots$ by choosing the $i$th term of the $i$th subsequence in the preceding construction we obtain a subsequence which converges to a point of $M$.

We have seen that the set $M$ is a compact, connected metric space $($i.e., $M$ is a *continuum*$)$. We now wish to show that $M$ is homeomorphic to $[0, 1]$. Let $$h(x) = \sum_{i \ge 1} {{x_i}\over{2^i}}.$$We observe that $h$ is one-to-one since if $x \neq y$ and $n$ is the least positive integer $j$ such that $x_j \neq y_j$ and $x_n > y_n$ then $x_i > y_i$ for each $i \ge n$. Moreover, $h$ throws $M$ onto $[0, 1]$ and it is not difficult to show that $h$ is continuous $($it is simply the distance from the point to the origin in the Hilbert cube$)$. So, $h$ is a homeomorphism.

A few remarks are in order. The construction of the set $M$ can be made using any function from $[0, 1]$ onto $[0, 1]$. That $M$ is compact can be seen to be a consequence of the continuity of the function $f$ since we used nothing specific about the original mapping other than its continuity to derive these properties. Of course, the construction of the homeomorphism of $M$ onto $[0, 1]$ was very specific to the choice of $f$.