Convex and conic hull, geometric interpretation $$\operatorname{conv}\,X=\left\{\sum_{i=1}^N \lambda_i x_i \,\Bigg\vert\, N\in\Bbb N,\, x_i\in X,\, \sum_{i=1}^N \lambda_i = 1,\lambda_i \geq 0\right\}$$
$$\operatorname{cone}\,X=\left\{\sum_{i=1}^N \lambda_i x_i | N\in \Bbb N, x_i \in X, \lambda_i \geq 0\right\}$$
I am having trouble interpreting the sets above, i.e. the convex and conic hull. Can someone shed some light on the sets above, based upon how they are written?

Convex hull geometrically:  The convex hull of a set, $X$, is the set of all points of $X$, and all points that lie on any line between any of the points in $X$.
Conic hull geometrically: The conic hull of a set, $X$, is the covex hull of $X\cup \{0\}$.
Is that all I should be reading from above?

$X$ is some subset of euclidean space $\Bbb R^n$
 A: (1) If $X$ is a some subset in $\mathbb{R}^n$,
$N=1$ : For $x\in X$, $1x\in {\rm conv}\ X$ so that $$ X\subset {\rm conv}\ X $$
$N=2$ : For $x,\ y\in X$, $$ \lambda_1x+\lambda_2 y \in {\rm conv}\ X,\ \lambda_1+ \lambda_2=1,\ \lambda_i\geq 0 $$  That is any line between points in $X$ is in ${\rm conv}\ X$. 
(2) And $Y:={\rm conv}\ X$ is convex : If $x,\ y\in Y$, then $$ x= \sum_{i=1}^k\lambda_i x_i,\ y=\sum_{j=1}^{l} \tau_j y_i,\ x_i,\ y_j\in X $$
Then for $a,\ b\geq 0,\ a+b=1$, then we have $$ ax+by = \sum_{i=1}^ka\lambda_i x_i + \sum_{j=1}^{l} b\tau_j y_i $$ And note that $$ 
 \sum_{i=1}^ka\lambda_i  + \sum_{j=1}^{l} b \tau_j =1 $$
(3) $Y$ is smallest set which is a convex set containing $X$ : If $Z$ is a convex set containing $X$, then any line between points in $X$ is in $Z$. And if $$\lambda_i\geq 0,\ \sum_{i=1}^3\lambda_i=1,\ x_i\in X $$ then $$  x:=\sum_{i=1}^3\lambda_i x_i=
 \lambda_1 x_1 + (1-\lambda_1) \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } x_i  $$ Note that $$ \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } =1 $$
So $ \sum_{i=2}^3 \frac{\lambda_i}{ 1-\lambda_1 } x_i \in Z$ so that $x\in Z$ Continuously we have $  \sum_{i=1}^k\lambda_i x_i\in Z$ so that ${\rm conv}\ X\subset Z$. 
(4) By definition $$ {\rm conv}\ X\subset {\rm cone}\ X $$
Let $$ Z:= \{ x\in \mathbb{R}^n\mid tx \in {\rm conv}\ X,\ {\rm
some}\ t\geq 0\} \cup \{0\}$$ Then
$$ Z = {\rm cone}\ X$$
Proof : $Z\subset {\rm cone}\ X $ : $$ x\in Z\rightarrow tx \in {\rm
conv}\ X \rightarrow tx=\sum_i c_i x_i,\ \sum_i c_i =1
$$
so that $$ x= \sum_i \frac{c_i}{t} x_i \in {\rm cone}\ X $$
${\rm cone}\ X\subset Z $ : Let $x=\sum_i c_i x_i \in {\rm cone}\
X,\ c:=\sum_i c_i$. Then $$ \frac{x}{c} = \sum_i \frac{ c_i}{c} x_i
\in {\rm conv}\ X $$ Let $t= \frac{1}{c}$ so that $x\in Z $.
A: The conic hull does more than just the convex hull of $X \cup \lbrace 0 \rbrace$. You could look at it as the convex hull of every ray from $0$ through points in $X$.
Also, the interpretation of the convex hull is incomplete. For example, the convex hull of the points $(0,0), (1,0), (0,1) \in \mathbb{R}^2$ is the filled-in triangle with those vertices. The point $(1/4, 1/4)$ lies in the convex hull as we may write,
$$(1/4, 1/4) = (1/2)(0, 0) + (1/4)(1, 0) + (1/4)(0, 1),$$
but $(1/4, 1/4)$ does not lie on any line segment between any two of the points $(0, 0), (1, 0), (0, 1)$.
The convex hull is not just every point on every line segment between every pair of points in $X$, but it is every point in every simplex we can form from a finite number of points in $X$.
(FYI, Caratheodory's theorem tells us that, in $\mathbb{R}^n$, we need only consider simplices of at most $n + 1$ points, so in $2$ dimensions, we need only consider triangles, and in $3$ dimensions, we need only consider tetrahedra.)
