(1) In each part, decide whether the statement is true or false. You do not need to justify your responses:
(a) If $f: X \rightarrow Y$ is an injective function then for all $y \in Y$ the set $f^{-1}(\{y\})$ has cardinality at most 1 .
(b) The relation $\sim$ on $\mathbb{Z}$ defined by $n \sim m$ if $n-m=1$ is an equivalence relation.
(c) If $A$ and $B$ are countable sets it is possible that $A \cup B$ is uncountable.
(d) If $X \subseteq \mathbb{Q}$ is nonempty and bounded above then $\sup (X) \in \mathbb{Q}$.
(e) The union of two or more open intervals is an open interval.

Step 1 ：(a) If $f: X \rightarrow Y$ is an injective function then for all $y \in Y$ the set $f^{-1}(\{y\})$ has cardinality at most 1. This statement is \(\boxed{\text{True}}\). An injective function, also known as a one-to-one function, is a function that never maps distinct elements of its domain to the same element of its codomain. Therefore, for each $y \in Y$, there can be at most one $x \in X$ such that $f(x) = y$. Hence, the preimage of any singleton set $\{y\}$ under $f$ has at most one element.

Step 2 ：(b) The relation $\sim$ on $\mathbb{Z}$ defined by $n \sim m$ if $n-m=1$ is an equivalence relation. This statement is \(\boxed{\text{False}}\). An equivalence relation on a set must satisfy three properties: reflexivity ($n \sim n$ for all $n$), symmetry ($n \sim m$ implies $m \sim n$), and transitivity ($n \sim m$ and $m \sim p$ implies $n \sim p$). The given relation is not symmetric, because if $n \sim m$ (i.e., $n - m = 1$), it does not follow that $m \sim n$.

Step 3 ：(c) If $A$ and $B$ are countable sets it is possible that $A \cup B$ is uncountable. This statement is \(\boxed{\text{False}}\). The union of two countable sets is also countable.

Step 4 ：(d) If $X \subseteq \mathbb{Q}$ is nonempty and bounded above then $\sup (X) \in \mathbb{Q}$. This statement is \(\boxed{\text{True}}\). The supremum of a set of rational numbers is also a rational number.

Step 5 ：(e) The union of two or more open intervals is an open interval. This statement is \(\boxed{\text{False}}\). The union of two or more open intervals is not necessarily an open interval. For example, the union of the intervals $(0, 1)$ and $(2, 3)$ is not an open interval.