Given $X = \{0\} \cup \{\frac{1}{n} |n>0, n\in Z \}$ and $Y =\{\frac{1}{n} |n>0, n\in Z \}$ with relative topology, I try to show that $H_q(X) \cong H_q(Y)$ and that $\pi_q(X,x_0)\cong\pi_q(Y,y_0)$. Furthermore that X and Y are not homotopy equivalent.
I have gotten that since $X\times I $ retracts to $X\times \{0\}\cup Y\times I$ the inclusion from Y to X is a cofibration. Would this help in showing the homotopy case?
On
The homology groups are given by the direct sum of the homology group of each path-connected component, and your path-connected components are points, hence it is easy to see that these two spaces have the same homology groups.
Similarly, $\pi_q(X,x_0) \cong \pi_q(C,x_0)$, where $C$ is the path-connected component containing $x_0$, hence you also have the same homotopy groups.
However, $X \times I$ doesn't retract to $X \times \{0\} \cup Y \times I$.
Given any totally path disconnected space $X$ we have that $H_n(X)=0$ for $n>0$ and $H_0(X)=\mathbb{Z}^{|X|}$. That's because the only continuous maps $\Delta^n\to X$ are constant maps.
Analogously $\pi_n(X)=0$ because a continuous map $S^n\to X$ has to be constant.
Therefore your $X$ and $Y$ have the same homology and homotopy groups, since both are totally path disconnected and equinumerous.
It is a bit harder to show that these are not homotopy equivalent. Assume that $f:X\to Y$ is a homotopy equivalence with its homotopy inverse $g:Y\to X$. Since $X$ is compact and $Y$ is discrete, then $im(f)$ is a finite subset of $Y$ and thus $im(g\circ f)$ is finite subset of $X$.
But in $X\times I$ path components are precisely $\{x_0\}\times I$. And so there is no homotopy $X\times I\to X$ between the identity and a non surjective map $X\to X$ (note that final image implies this). Because of vertical paths: given $x_0\not\in im(g\circ f)$ we can only connect $(x_0,1)$ to $(x_0,0)$, no other $(y,0)$ lies in the path component of $(x_0,1)$. Contradiction.