Question asking relation between vertex cover and independent set Question

$\text{A vertex cover in an undirected graph G is a subset } C⊆V(G) \text{such that every edge of G}$
$\text{has an endpoint in C. An independent set in G is a subset }I⊆V(G)$
$\text{such that no edge has both its endpoints in I.}$

Which of the following is TRUE of every graph $G$ and every vertex cover $C$ of $G?$
$1.\text{There exists an independent set of size ∣C∣}$
$2.V(G)−C  \text{is an independent set}$
$3.∣C∣≥\frac{∣E(G)∣}{2}$
$4.∣C∣≥\frac{∣V(G)∣}{2}$
This question was asked in one of the entrance examination.As i am practicing some good graph questions, i willed to solve this.
My Approach
I solve it by taking an example graph ,I took a cycle graph $C_5$.
$ABCDEA$ is my cycle graph.
$I=\left \{  A,C\right \}$
OR
$I=\left \{  B,D,E\right \}$

i have little doubt in recognising $C$,
I think $C=V=\left \{  A,B,C,D,E\right \}$
i.e i think $C$ will always be vertex set.
so based on above analysis,
$|C|=5,|I|=3 \text{or} |I|=2$

1.There does not exists an Independent set of size $5$.
2.$V(G)-C $ is independent set
I am lost in other option .Please help me out.
 A: *

*For the graph $G = C_3$ with $V(C_3) = \{x_1, x_2, x_3\}$, $|C| = 2$ and $|I| = 1$ for all the cases because $C = \{x_1, x_2\}$ or $\{x_1, x_3\}$ or $\{x_2, x_3\}$; and $I = \{x_1\}$ or $\{x_2\}$ or $\{x_3\}$. So the argument is FALSE because every graph $G$ does not satisfy the given equality.

*I think this is TRUE because of the definitions. Since $C$ includes at least one end point of each edge, $V(G) - C$ must include either none of the endpoints of one edge or 1 of them. So it is an independent set by definition.

*For this one, I can give a counterexample. Consider the graph $G(V,E)$ with $V = \{x_0, x_1, x_2, x_3\}$ and $E = \{\{x_0, x_1\}, \{x_0, x_2\}, \{x_0, x_3\}\}$ that is a minimal example of a tree(but also a graph) with 3 leaves as shown here. In this tree, notice the set $\{x_0\}$ is one of the vertex covers. So we can find a vertex cover $C$ with $|C| = 1$, where $|C| = 1 < 3/2 = |E(G)/2|$. So the argument is FALSE because every vertex cover of given graph $G$ does not satisfy the given inequality.

*Same counterexample in part 3 can be given for this case. Only difference is the last inequality, which is for this case, $|C| = 1 < 2 = |V(G)/2|$.

