Show that, for every non-empty subset A of [ n ] , there is an intersecting family F of subsets of [ n ] of size $2^{n − 1}$ with A ∈ F .

Here is what I have understood this question as:

Every subset of [n] is contained in an intersecting family of size 2^n-1

This is my approach:

I took a set, X={1,2,3}

I list down all the subsets of X: {}, {1}, {2}, {3}, {1,2}, {2,3}, {1,3 }, {1,2,3}

Now, I take any one from the above, let us take {2,3}, and name it A

To build an Intersecting family containing A, I take any element from A, let us say, 3. I remove 3 from X and list down all the subsets of our new X: {}, {1}, {2}, {1,2}. I further add 3 to each of these sets, so I have now: {3}, {1,3}, {2,3}, {1,2,3}.

The above list forms an intersecting set containing A, and is of size 4(2^3-1).

In the same way, I can repeat the above steps for all subsets of X, and conclude the same solution.

I apologise for a naive question, but I would like to know if I am correct in this approach, and If I am, How should I approach for a mathematical proof?

Welcome to this site. Is the following all you require or have I misunderstood your question?

Let $$x$$ be an element of a set $$X$$ which has $$n$$ elements.

The set $$X-\{x\}$$ has $$n-1$$ elements and therefore has $$2^{n-1}$$ subsets. Add $$x$$ back into each of these subsets to give us the required family.

• Thankyou for your warm welcome. Absolutely, Once again, thank you. Sep 10 '19 at 8:47

Your arguing about the set $$X$$ is correct, albeit somewhat laborious. But $$X=\{1,2,3\}$$ is just a special case. Now you have to extend your argument to the general setup in the question; then erase the proof concerning $$X$$.

You are given a nonempty subset $$A\subset[n]:=\{1,2,\ldots, n\}$$. Choose a number $$a\in A$$ and consider the family $${\cal F}:=\bigl\{X\subset[n]\bigm| a\in X\bigr\}\ .$$

• Thankyou Christian, It solves my problem very efficiently Sep 10 '19 at 9:02