# Element of Toset has at most One Immediate Predecessor

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## Theorem

Let $\struct {S, \preceq}$ be a toset.

Let $a \in S$.

Then $a$ has at most one immediate predecessor.

## Proof

Let $b, b' \in S$ be immediate predecessors of $a$.

We have that $\preceq$ is a total ordering.

- $b \preceq b'$

By virtue of $b$ being a immediate predecessor of $a$:

- $\neg \exists c \in S: b \prec c \prec a$

However, since $b'$ is also an immediate predecessor:

- $b' \prec a$

Hence, it cannot be the case that $b \prec b'$.

Since $b \preceq b'$, it follows that $b = b'$.

Hence the result.

$\blacksquare$

## Also see

## Sources

- 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.11$