Proof of a Corollary to Hall's Marriage Theorem I'm trying to prove Corollary 3.3 from this paper (http://www.sfu.ca/~mdevos/notes/graph/345_matchings.pdf) and am lost at the step that says: "Similarly, every vertex in N(X) has degree k, so t is less than or equal to k|N(X)|. It follows that |X| is at most |N(X)|."
How do we know that t is less than or equal to k|N(X)|?
Thanks so much!

Every vertex in $X$ has at least $k$-neighbors (that are all in $Y$ because the graph is bipartite
So if $|X| \le k$ we are done since $|N(X)| \ge k$
If $|X|=n>k$ then there is at least $n$ vertices in $N(X)$ since if there is less than $n$ vertices then you'll find a vertex in $N(X)$ of degree $> k$, a contradiction.
Thus, $|X| \le |N(X)|$
