Homology of the closed topologist sine curve The closed topologist sine curve, $X$, is the subspace of $R^2$ consisting of all the points $(x,\sin(1/x))$ for $x \in (0,1]$, and all points $(0,y)$ for $y \in [-1,1]$ and an arc from $(0,-1)$ to $(1,\sin(1))$ Compute the singular homology of $X$, using a suitable Mayer-Vietoris sequence.
So to utilize the Mayer-Vietoris sequence I need an excisive couple, one condition that will gaurentee that a pair is an excisive pair is if the union of their interiors cover the space. 
$X_1$ = $(x,\sin(1/x))$ for $x \in (0,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$ 
$X_2$ = all points $(0,y)$ for $y \in [-1,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$
I don't know i'm just taking a shot in the dark really I guess. Can anyone offer some insight for me? Thanks!!

We do not need the Mayer-Vietoris sequence and what we use is primary.
Computing the homology group of the part $X_2=\lbrace(0, y)\ |\ -1 \leq y \leq 1\rbrace$ is easy because it is bounded and convex. Hence $n=0$，$H_0(X_2)=Z$ and $n \geq 1$，$H_n(X_2)=0$
The part of $X_1=\lbrace(x, \sin(\frac{1}{x})\ |\ 0 < x \leq 1\rbrace$ is a bit complicated. Define $f \in S_n(X_1)$ and $f(t)=(x(t), y(t))$ where $t$ is a point of $n$-simplex. For the $n$-simplex is closed, $\inf x(t) > u > 0$. Hence we can define $K=\lbrace(x, \sin(\frac{1}{x}))\ |\ u \leq x \leq 1\rbrace$, then $im(f) \subseteq K$.
If $cls\ y \in H_n(X_1)$, we have $supp\ y \subseteq \text{some}\ K$. And assume $j$ is the inclusion between $K$ and $X_1$. Then $cls\ y \in im\ j^*$ where $j^*$ is the inclusion between $H_n(K)$ and $H_n(X_1)$. Because $K$ is homeomorphic to $I=[0, 1]$, $n \geq 1$, $H_n(K)=0$. Hence $cls\ y = 0$ in $H_n(X_1)$ and $n \geq 1$, $H_n(X_1)=0$.
Then finally $n=0$, $H_0(X)=Z \oplus Z$ and $n \geq 1$, $H_n(X)=0$.
