I am trying to make sure my intuition for the following question from an assignment is correct

Prove or disrove: if $G = (V, E)$ is a graph and $I_1$ and $I_2$ are independent sets in $G$, then $I_1 \cap I_2$ is an independent set in $G$.

Answer: This statement is true. Consider arbitrary nodes $x, y \in I_1$ and $u, v \in I _2$. Since $I_1$ and $I_2$ are independent sets in $G$, there is no edge between $\{x, y\}$ and there is no edge between $\{ v, u \}$. Similarly, here is no edge between $\{x, u\}$ and $\{v, y\}$ because otherwise, $I_1$ and $I_2$ would have to have nodes that are adjacent two each other. We see that $I_1 \cap I_2 = \emptyset$ and therefore, the statement that $I_1 \cap I_2$ is an independent set in $G$ is vacuously true.

Is this approah necessarily true for any two independent sets in a graph? I would appreciate any feedback

  • $\begingroup$ We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. $\endgroup$ Commented Feb 11, 2018 at 13:16
  • $\begingroup$ But, by way of feedback, it's not true that there can be no edges between independent sets and your claim that $I_\cap I_2=\emptyset$ doesn't follow from anything you've written, although it is true that $\emptyset$ is vacuously independent $\endgroup$ Commented Feb 11, 2018 at 13:19

1 Answer 1


Your proof is unfortunately wrong. Consider what happens when $I_1 = I_2$, for example.

The essential reason behind the statement is the following more general fact:

Any subset of an independent set is an independent set.

You can try proving this rigorously.

  • $\begingroup$ so can I assume (or define $I_1 = I_2$ in my proof? But then this would not be "any" two independent sets but just these two specific ones? I am not sure how to proceed? $\endgroup$
    – user75706
    Commented Feb 10, 2018 at 21:55
  • $\begingroup$ Is there a similar reasoning behind this statement as well? : Prove or disprove: if $G = (V, E)$ is a graph and $C_1$ and $C_2$ are vertex covers of $G$, then $C_1 \cap C_2$ is a vertex cover of $G$ $\endgroup$
    – user75706
    Commented Feb 10, 2018 at 22:02
  • $\begingroup$ Vertex covers satisfy the dual property: a superset of a vertex cover is a vertex cover. $\endgroup$ Commented Feb 10, 2018 at 22:04

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