A set of edges $I$ of a graph $G(V, E)$ is independent if $I$ contains no cut set of G. If $I$ and $J$ are independent, then prove there exists an edge $e$ in $J \setminus I$ such that $I \cup{e}$ is independent.
The graph $G(V, E)$ is finite and a cut set $S \subset E$ is a set such that the graph $G(V, E \setminus S)$ is disconnected.
My attempt
Case 1: $I \subset J$
Then any $e \in J \setminus I$ will do.
Case 2: $I \not\subset J$ but $I \cap J \ne \varnothing$
Let $e$ be an edge in $I \cap J$. If there is an edge, $x$, adjacent to $e$ in $J \setminus I$, then $I \cup \{x\}$ is independent.
I don't know how to proceed if there is no such edge $x$. Nor do I have any idea how to handle the third case where $I$ and $J$ are disjoint.
Throughout this answer, a "graph" is always assumed to be finite.
Definition. A bridge of a connected graph $(V,E)$ is an edge $e \in E$ such that $\{e\}$ is a cutset, i.e. $(V,E \setminus \{e\})$ is disconnected.
Lemma. Given a connected graph $(\tilde{V},\tilde{E})$ and a set $F \subset \tilde{E}$, the number of connected components of $(\tilde{V},\tilde{E} \setminus F)$ is at most $|F|+1$. If every element of $F$ is a bridge of $(\tilde{V},\tilde{E})$ then the number of connected components of $(\tilde{V},\tilde{E} \setminus F)$ is exactly $|F|+1$.$\ $ [Proof given below.]
Now suppose the statement is false, i.e. we have a graph $(V,E)$ and sets $I,J \subset E$ such that $|J|>|I|$, the graphs $(V,E \setminus I)$ and $(V,E \setminus J)$ are connected, and every element of $J \setminus I$ is a bridge of $(V,E \setminus I)$. On the one hand, applying the Lemma to the graph $(\tilde{V},\tilde{E})=(V,E \setminus I)$ implies that the number of connected components of $(V,E \setminus (I \cup J))$ is equal to $|J \setminus I|+1$. But on the other hand, applying the Lemma to the graph $(\tilde{V},\tilde{E})=(V,E \setminus J)$ implies that the number of connected components of $(V,E \setminus (I \cup J))$ is at most $|I \setminus J|+1$. Thus we have that $|J \setminus I| \leq |I \setminus J|$, which contradicts our assumption that $|J|>|I|$.
Proof of the Lemma:
To prove the first part, it is sufficient to show that the removal of one edge $e$ from any graph $(V,E)$ increases the number of connected components by at most one. Letting $\mathcal{C}$ be the set of connected components of $(V,E \setminus \{e\})$, we have that either (i) the two vertices of $e$ belong to the same member of $\mathcal{C}$, in which case the set of connected components of $(V,E)$ is $\mathcal{C}$, or (ii) the two vertices of $e$ belong to different members $C_1$ and $C_2$ of $\mathcal{C}$, in which case the set of connected components of $(V,E)$ is $\,\mathcal{C} \cup \{C_1 \cup C_2\} \setminus \{C_1,C_2\}$.
For the second part, let $\tilde{\mathcal{C}}$ be the set of connected components of $(\tilde{V},\tilde{E} \setminus F)$, and equip $\tilde{\mathcal{C}}$ with a graph structure $\mathcal{E}$ defined by $\{C_1,C_2\} \in \mathcal{E}$ if and only if there is an edge in $F$ with one vertex in $C_1$ and the other in $C_2$. We show that $(\tilde{\mathcal{C}},\mathcal{E})$ contains no cycle, from which it follows that $|\tilde{\mathcal{C}}|>|\mathcal{E}|$ (Prove by induction that every graph with n vertices and at least n edges has a cycle).
Suppose for a contradiction that we have $n \geq 1$, distinct sets $C_1,\ldots,C_n \in \tilde{\mathcal{C}}$ and $a_1,b_1,\ldots,a_n,b_n \in \tilde{V}$ with $\{a_i,b_i\} \in F$ for each $1 \leq i \leq n$, such that $a_i \in C_i$ for each $1 \leq i \leq n$, $\,b_i \in C_{i+1}$ for each $1 \leq i < n$, and $b_n \in C_1$. Let $e=\{a_n,b_n\}$; we want to obtain a contradiction by showing that $(\tilde{V},\tilde{E} \setminus \{e\})$ is connected. Since $(\tilde{V},\tilde{E})$ is connected, it is sufficient to show that there is a path in $\tilde{E} \setminus \{e\}$ joining $b_n$ to $a_n$. Since $b_n,a_1 \in C_1$, there is a path in $\tilde{E} \setminus F \,\subset\, \tilde{E} \setminus \{e\}\,$ joining $b_n$ and $a_1$; likewise there is a path in $\tilde{E} \setminus F$ joining $b_i$ to $a_{i+1}$ for each $1 \leq i < n$. And for each $1 \leq i < n$, the edge $\{a_i,b_i\}$ is distinct from $e$. Thus we have constructed a path in $\tilde{E} \setminus \{e\}$ from $b_n$ to $a_n$.
So we have shown that $(\tilde{\mathcal{C}},\mathcal{E})$ contains no cycle and so $|\tilde{\mathcal{C}}|>|\mathcal{E}|$. But also, as a special case of the above argument, we have that for any $C_1,C_2 \in \tilde{\mathcal{C}}$ there is at most one edge in $F$ having one vertex in $C_1$ and the other in $C_2$. Thus we conclude that $|\tilde{\mathcal{C}}|>|F|$, and so by the first part of the lemma, $|\tilde{\mathcal{C}}|=|F|+1$.
Now some terminology clarifications.
Firstly, this seems to be a very unusual use of the word "independent". Usually, a set of edges $C \subset E$ is called independent if no two edges in $C$ share a vertex. In this case, the statement becomes false, as illustrated just by taking $J=\{\{x_0,x_1\},\{x_2,x_3\}\}$ and $I=\{\{x_1,x_2\}\}$; but it can be corrected simply by strengthening the requirement $|J|>|I|$ to $|J|>2|I|$. This is due to the fact that the total number of vertices contained in members of $I$ is $2|I|$ and none of these vertices can belong to more than one edge in $J$.
Secondly, I notice (from Googling) that there appear to be different definitions of a cutset [here meaning an "edge cutset"], but they should still give rise to the same notion of "containing a cutset". To be precise: Given a finite graph $(V,E)$, three possible definitions of a cutset, listed in order of increasing strength [I think!], are:
Since I assume that the graph is finite, it is easy to see that every "definition 1 cutset" contains a "definition 3 cutset". Therefore, if I am right that the three above definitions are in order of increasing strength, then it immediately follows that the statement "$I$ contains a cutset" is equivalent to the statement that $(V,E \setminus I)$ is disconnected, under any of the three definitions of cutset.