Quasi circle is not contractible I'm trying to show that the quasi circle (picture below) doesn't have the homotopy type of a CW complex. I proved that all homotopy groups are zero. Now I need to show that it is not contractible to use Whitehead's theorem. I know that we can collapse the vertical interval to a point to get a circle. Maybe we can use this map to show that the quasi circle is not contractible, but I don't know how.
Thanks

 A: Say we want to contract it to $(0, 0)$. Look at a point on the vertical interval. Any neighbourhood of that point contains some points of the sine-curve, but any point on the sine curve has to go all the way back through $(1, 0)$ to get to the origin. Thus a contraction is impossible to do continuously.
A: The easiest argument is to observe that the 1st Cech cohomology of this space is nontrivial. You can see this by observing that it separates the plane in 2 components and applying Alexander duality. 
A: I wanted to provide a detailed argument that doesn't use any technology. First, notation: Let $x$ denote $(0,0)$, and suppose we have a homotopy $F:X \times I \rightarrow X$ which is the identity for $t=0$ and the constant map to $x$ for $t = 1$. Denote the vertical line segment of bad cluster points by $L$, and let $U$ denote an open set containing the sine curve, say with $0 < x < 1$. Let $x_n = (\frac{1}{n},0)$, so $x_n \rightarrow x$. 
Let $S$ be the set of times $t$ such that $F(x,t) \in L$ and $F(x_n,t) \in U$ for large  $n$. I'll show this set is open and closed, so we win. 
For closedness, pick a sequence $t_i$ of times in $S$ converging to $t$. Since $L$ is closed, we have to have $F(s,t) \in L$, so $t$ satisfies our first condition. Select an open set $V$ about $F(x,t)$ so that the various line segments of the sine curve cannot join one another without leaving $V$. There is then a time interval $(t-\epsilon,t+\epsilon)$ and an open set $W$ about $p$ for which $(q,s) \in W \times (t - \epsilon, t+\epsilon)$ must be sent into $V$. Pick any $t_n$ which falls inside $(t - \epsilon, t+\epsilon)$, and suppose WLOG that $t_n < t$. Then for large $m$, we know that $F(x_m,t_n) \in V$. But by the choice of $V$, the path defined by $F(x_m,\cdot)$ as we vary from $t_n$ to $t$ cannot leave $V \cap U$. So for large $m$, $F(x_m,t) \in U$.
For openness, let $V, \epsilon, W$ be as above. Arguing in a similar manner, we have that for each $s \in (t-\epsilon, t+\epsilon)$, the points $F(x_n,s)$ are trapped in $U \cap V$. On the other hand, we know that $F(x_n,s) \rightarrow F(x,s)$, and so we must have $F(x,s) \in L$, being the limit point of points on the sine curve.
This proof can probably be adapted to lots of other situations with some kind of local nonconnectedness.
