Prove that a numerical semigroup $S$ is symmetric if and only if it equals its canonical ideal. Recall that a numerical semigroup $S$ is a submonoid of the positive integers such that $\mathbb Z_{\geq 0} \setminus S$ is finite. Consequently, the largest positive integer not contained in $S$ is well-defined; it is known as the Frobenius number $\operatorname F(S).$ Put another way, we have that $$\operatorname F(S) = \max \{n \mid n \in \mathbb Z_{\geq 0} \setminus S\}.$$ Further, recall that the collection of pseudo-Frobenius numbers of $S$ is defined as $$\operatorname{PF}(S) = \{n \in \mathbb Z_{\geq 0} \setminus S \mid n + s \in S \text{ for all nonzero elements } s \in S\}.$$ Observe that $\operatorname F(S)$ must be pseudo-Frobenius; otherwise, there would be a larger integer not contained in $S.$ We say that $S$ is symmetric if and only if $\operatorname{PF}(S) = \{\operatorname F(S)\}.$
We say that a set $I \subseteq \mathbb Z$ is a relative ideal of $S$ whenever $(1.)$ $I + S \subseteq I$ and $(2.)$ there exists an element $s \in S$ such that $s + I \subseteq S.$ The canonical ideal of $S$ is the relative ideal $$\Omega = \{n \in \mathbb Z \mid \operatorname F(S) - n \notin S\}.$$

(Exercise 4.22, Numerical Semigroups by Rosales & García-Sánchez) We have that $S$ is symmetric if and only if $\Omega = S.$

One direction of the proof is clear to me. For if $\Omega = S,$ then every integer $s$ satisfying $\operatorname F(S) - s \notin S$ must belong to $S.$ In particular, every pseudo-Frobenius number can be written as $\operatorname F(S) - s$ for some positive integer $s.$ By definition, the integer $\operatorname F(S) - s$ does not belong to $S,$ so $s$ must belong to $S,$ and therefore, $\operatorname F(S) = [\operatorname F(S) - s] + s$ must belong to $S$ — a contradiction. But the other direction — namely, if $\operatorname F(S)$ is the only pseudo-Frobenius number, then $\Omega = S$ — is giving me some trouble. Perhaps, it is possible to prove this by contradiction, but I am unravelling the definitions, and nothing is coming to mind.
I would greatly appreciate any advice, comments, or suggestions. Thank you for your time.
 A: Let $S\subset \mathbb{Z}_{\geq 0}\;$(proper inclusion) be a numerical semigroup.

For convenience of notation, let $P=\operatorname{PF}(S)$, and let $f=\operatorname{F}(S)$.

Claim:$\;$If$\;P=\{f\}$, then $\Omega=S$.

Proof:

Assume $P=\{f\}$.

First we show $S\subseteq\Omega$.

Let $s\in S$.

If $f-s\in S$, we would have $f-s=s'$ for some $s'\in S$, but then $f=s+s'$ would yield $f\in S$, contradiction.

Hence $f-s\not\in S$, so $s\in\Omega$.

Therefore $S\subseteq\Omega$.

Hence to show $\Omega=S$, it will suffice to show $\Omega{\large{\setminus}}S={\large{\varnothing}}$

Suppose instead that $\Omega{\large{\setminus}}S$ is nonempty.

Our goal is to derive a contradiction.

Since $S$ contains all positive integers greater than $f$, it follows that $\Omega{\large{\setminus}}S$ has a largest element, $x$ say.

By definition of $f$, it follows that all elements of $\Omega$ are nonnegative.

Since $0\in S$, it follows that $f\not\in\Omega$, hence $0 < x < f$.

Let $s\in S$ with $s > 0$.$\;$Then

\begin{align*}
&
x\in\Omega
\\[4pt]
\implies\;&
x+s\in\Omega
&&\bigl(\text{since $\Omega$ is a relative ideal}\bigr)
\\[4pt]
\implies\;&
x+s\in S
&&\bigl(\text{by maximality of $x$}\bigr)
\\[4pt]
\implies\;&
x\in P
&&\bigl(\text{since $s$ is an arbitrary positive element of $S$}\bigr)
\\[4pt]
\implies\;&
x=f
\\[4pt]
\end{align*}
contradiction, which completes the proof.
