What is $H_0(X)$ where $X$ is a space consisting of k distinct disconnected points? This might be a dumb question but I would just want to clarify my doubt.
Is it correct if I say the zeroth singular homology group of k distinct points is $\mathbb{Z}^k$? 
Is it because the zeroth homology tells about how many disconnected components in a space?
So $H_0(X)=\frac{\ker\partial_0}{\text{Im }\partial_1}=\frac{\mathbb{Z}^k}{\{0\}}=\mathbb{Z}^k$.
If I am wrong, what are my mistakes? Could somebody please clarify this? Thanks.
 A: Yes, this is correct. Write $X = \{ x_1, \dots, x_k \}$.
The group $C_0(X)$ is the free abelian group on the set of continuous maps $\Delta^0 \to X$, where $\Delta^0$ is a one-point space. It's obvious here that there's a generator $x_i : \Delta^0 \to X$ for each $i$ given by $x_i(*) = x_i$, so $C_0(X) \cong \mathbb{Z}^k$.
The group $C_1(X)$ is the free abelian group on the set of continuous maps $\Delta^1 \to X$, where $\Delta^1 = [0,1]$ is an interval. It's also clear that such a map has to be constant (because $X$ is discrete), so you get a generator $g_i : [0,1] \to X$ given by $g_i(t) = x_i$ for each $i$, and $C_1(X) \cong \mathbb{Z}^k$ too.
Now, $\partial_0 : C_0(X) \to C_{-1}(X)$ is of course zero, so $\ker \partial_0 = \mathbb{Z}^k$. Moreover, when we compute $\partial_1 : C_1(X) \to C_0(X)$, we find that $\partial(g_i) = g_i(1) - g_i(0) = x_i - x_i = 0$. Thus $\operatorname{im} \partial_1 = 0$, and finally $H_0(X) \cong \mathbb{Z}^k/0 = \mathbb{Z}^k$.
More conceptually, it's well-known (and not hard to prove) that the zeroth homology group of a singleton is $\mathbb{Z}$, and the homology of a disjoint union is given by $H_0(X \sqcup X') = H_0(X) \oplus H_0(X')$. So in your case you get
$$H_0(X) = H_0(\{x_1\}) \oplus \dots \oplus H_0(\{x_k\}) = \mathbb{Z} \oplus \dots \oplus \mathbb{Z} = \mathbb{Z}^k.$$
