Algebraic and homological equivalence relations for $0$-cycles Let $X$ be a connected smooth projective variety. Let  $Z_0(X)_{alg}$ be the group of $0$-cycles algebraically equivalent to $0$ and $Z_0(X)_{\hom}$ be the group of $0$-cycles homologically equivalent to $0$.
In Voisin's book "C.Voisin. Hodge Theory and Complex Algebraic Geometry,  II", section 8.2.1 says:
a $0$-cycle is always supported on a smooth connected curve contained in $X$,
and since the cycles homologous to $0$ are the $0$-cycles of degree $0$, we
conclude that
\begin{equation*}
     Z_0(X)_{alg}= Z_0(X)_{\hom}.
\end{equation*}
I do not understand very well the statement to justify  the above equality. Could someone give a more detailed explanation or the  intuition behind this explanation so this equality holds. Thank you  very much!
 A: This is a Bertini's theorem-type argument. It suffices to show
Claim: For a given finite set $S\subset X$, there exists a smooth hypersurface $X'$ of $X$ containing $S$.
Then by induction on the dimension, one finds a smooth curve $C$ in $X$ containing $S$. In particular, for any $\alpha\in Z_0(X)_{hom}$, one can choose the homologous relation $\alpha\sim0$ to be supported on a smooth curve (drag negative points to positive points along real paths). Therefore $\alpha$ is algebraically trivial as well.
To prove the claim, let $\mathbb P^N$ be the ambient projective space containing $X$. Let $S=\{p_1,\ldots,p_k\}$. First, for each $p_i$, we can find a hyperplane $H_i$ containing $p_i$ and intersects transversely with $X$. ($H_i$ corresponds to a point in the hyperplane $p_i^{\vee}$ of the dual space $(\mathbb P^N)^{\vee}$ but not on the dual variety $X^{\vee}$.)
Now since the number of points $k$ could be larger than the dimension of ambient projective space, we should consider a linear system $|\mathcal{O}_{\mathbb P^N}(H_1\cup\cdots\cup H_k\cup nH_0)|=|\mathcal{O}_{\mathbb P^N}(k+n)|$, where $n$ is some large number (at least $N-k+1$), and $H_0$ is a general hyperplane. In particular, the sub-linear system $\mathcal{D}$ of $|\mathcal{O}_{\mathbb P^N}(k+n)|$ containing $S$ is not empty, and by Bertini's theorem, a generic element $Y\in \mathcal{D}$ intersects transversely with $X$ away from $S$. But by choosing an element $Y$ sufficiently close to $H_1\cup\cdots\cup H_k\cup nH_0$ (in analytic topology), $Y$ intersects $X$ at each $p_i$. So $Y\cap X$ is smooth and contains the finite set $S$.
