Suppose that we have two arbitrary sets $\mathcal{X}$, $\mathcal{Y}$ and are given a function $c : \mathcal{X} \times \mathcal{Y} \rightarrow \mathbb{R}$

Consider the following inequality for arbitrary points $(x_i, y_i) \in \mathcal{X} \times \mathcal{Y}$: $$\sum_{i=1}^N c(x_i, y_i) \leq \sum_{i=1}^N c(x_i, y_{i+1}) \quad \quad \textbf{(1)} \\ \text{ and denote } \Gamma_N \equiv \{((x_1, y_1), ..., (x_N, y_N)) \in (\mathcal{X} \times \mathcal{Y}) ^N \text{ such that } \textbf{(1)} \text{ holds}\}$$

We have the convention here that $y_{N+1} \equiv y_1$. Define proj$_k(((x_1, y_1), ..., (x_N, y_N))) = (x_k, y_k)$ and $$\Gamma \equiv \bigcap_{N \in \mathbb{N}}\bigcap_{1 \leq k \leq N} \text{proj}_k(\Gamma_N)$$

Then I claim for any $N \in \mathbb{N}$ and any collection $(x_1,y_1), ..., (x_N, y_N) \in \Gamma$, **(1)** holds. I DO NOT KNOW HOW TO SHOW THIS PROPERLY EVEN THOUGH IT'S APPARENTLY SUPPOSED TO BE COMPLETELY "OBVIOUS", so maybe I'm being dumb. I recognize that it's intuitive, but I can't write down a rigorous proof for it and I'm in desperate need of help because I've wasted a stupid amount of time on this.

This is all part of the proof of Kantorovich duality in Villani's 2009 book on optimal transport but my question is genuinely just about the sets/definition of projection. I've posted a screenshot below of the part of the book this is taken from.

**My attempt at showing this is as follows. I'm just writing out the definitions essentially because I'm lost:**

Suppose we have $(x_i, y_i) \in \Gamma, i \in \{1, ..., N\}$. This means that for every $i$, $(x_i, y_i) = \text{proj}_k (z)$ where $z \in \Gamma_N$. i.e. there exists a set of $N-1$ points such that **(1)** holds for any permutation of the $\mathcal{Y}$ indexed points.

I have absolutely no idea how $$\sum_{i=1}^N c(x_i, y_i) > \sum_{i=1}^N c(x_i, y_{i+1})$$ might produce a contradiction. Please help me if you can. This is probably so obvious and I recognize I'm too braindead to be studying math at this point.

You don't really even need to look at the screenshot but I'm just adding it for reference.