Proof of Lemma (Cardinality) - Tao's Analysis I I am having trouble understanding this proof from Tao's book:
Lemma 3.6.9. Suppose that $n \geq 1$, and $X$ has cardinality $n$. Then $X$ is non-empty, and if $x$ is any element of $X$, then the set $X \setminus \{ x \}$ (i.e., $X$ with the element $x$ removed) has cardinality $n−1$. 
Proof. If $X$ is empty then it clearly cannot have the same
cardinality as the non-empty set $\{i \in \mathbb{N} : 1 \leq i \leq n\}$, as
there is no bijection from the empty set to a non-empty set
(why?). Now let $x$ be an element of $X$. Since $X$ has the same
cardinality as $\{i \in \mathbb{N} : 1 \leq i \leq n \}$, we thus have a
bijection $f$ from $X$ to $\{i \in \mathbb{N} : 1 \leq i \leq n \}$. In
particular, $f(x)$ is a natural number between $1$ and $n$. Now deﬁne the
function $g : X \setminus \{x\}$ to $\{i \in \mathbb{N} :1 \leq i \leq n−1\}$ by the following
rule: for any $y \in X \setminus \{x\}$, we deﬁne $g(y) :=f(y)$ if $f(y) < f (x)$,
and deﬁne $g(y) :=f(y) − 1$ if $f(y) > f(x)$. (Note that $f(y)$ cannot equal
$f(x)$ since $y \neq x$ and f is a bijection.) It is easy to check that
this map is also a bijection (why?), and so $X \setminus \{x\}$ has equal
cardinality with $\{i \in N :1 \leq i \leq n−1\}$. In particular $X\setminus \{x\}$
has cardinality $n−1$, as desired.
Thoughts:
$X \setminus \{ x \} \subseteq X$ so $f$ is defined for every $y \in X \setminus \{ x \}$
so I can see why I can define $g(y):=f(y)$. I also know that $\{i \in N :1 \leq  i  \leq  n−1\}$ has one less element than $\{i \in N :1 \leq  i  \leq  n\}$. So if I match every $g(y)$ to every $f(y)$ there would be one element left from the second set which is $f(x)$ with no correspondence.
What I "feel" that Tao is trying to do is leaving the last element of $\{i \in N :1 \leq  i  \leq  n\}$ (which would be $n$) without correspondence. 
First time asking here so would be glad to know if I make a mistake in the way I am asking this question so I can avoid it next time.
Thanks.
 A: Remember that a finite set $S$ has cardinality $n$ if there exists a bijection $f : S \rightarrow \{ i \in \mathbb{N} \mid 1 \leq i \leq n \}$.
So to show that $X \setminus \{ x \}$ has cardinality $n-1$, we show that there exists a bijection $$g : X \setminus \{ x \} \rightarrow \{ i \in \mathbb{N} \mid 1 \leq i \leq n-1 \}.$$ We already know that $X$ has cardinality $n$, so there exists a bijection $$f : X \rightarrow \{ i \in \mathbb{N} \mid 1 \leq i \leq n \}. $$ Think of $f(a)$ (where $a \in X$) as the "index" of $a$.
We simply modify $f$ to not mention $x$; the resulting function $g$ has domain $X \setminus \{ x \}$ and range $\{ i \in \mathbb{N} \mid 1 \leq i \leq n-1 \}$. We do not "leave an element without correspondence".
Maybe an example would be helpful here.
Let $X = \{ a,b,c,d \}$. There is a bijection $f : X \rightarrow \{ i \in \mathbb{N} \mid 1 \leq i \leq 4 \}$ given by 
$$f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4$$
The "index" of $c$ is therefore $3$. Now the set $X \setminus \{c\}$ has cardinality $4-1 = 3$. Our new bijection $g$ agrees with $f$, except that all elements that have index greater than that of $c$ now have index one less. So our new bijection $g : \{ a,b,d \} \rightarrow \{ 1,2,3\}$ is 
$$g(a) = 1, g(b) = 2, g(d) = 3$$.
