Q.Determine whether the following statement is true or false. Justify your answer: For all sets $A$, $B$ and $C$, if $A \subset B$, then $A \cup C \subset B \cup C$.

The key idea is to prove the subset relationship by showing element-wise inclusion.

To determine if $A \cup C \subset B \cup C$ given $A \subset B$, we assume an arbitrary element $x$ belongs to $A \cup C$ and show it must also belong to $B \cup C$.

1. Assume $A \subset B$. This means that if $x \in A$, then $x \in B$.
2. Let $x$ be an arbitrary element such that $x \in A \cup C$.
3. By the definition of union, $x \in A$ or $x \in C$.
4. If $x \in A$, then because $A \subset B$, we have $x \in B$. Consequently, $x \in B \cup C$. …

The problem asks us to determine the truth value of a statement concerning set operations and subset relationships. Specifically, it states: "For all sets $A$, $B$ and $C$, if $A \subset B$, then $A \cup C \subset B \cup C$." We need to justify our answer.

Concept and Intuition

At its core, this problem tests our understanding of what "subset" and "union" mean.

• Subset ($X \subset Y$): This means that every element of set $X$ is also an an element of set $Y$. There are no elements in $X$ that are not in $Y$.
• Union ($X \cup Y$): This set contains all elements that are in $X$, or in $Y$, or in both.

The statement suggests that the subset relationship is "preserved" under the union operation. Imagine $A$ as a smaller collection of items completely contained within a larger collection $B$. Now, if we introduce a third collection $C$ and combine it with $A$ (to form $A \cup C$) and also combine it with $B$ (to form $B \cup C$), it intuitively makes sense that the "expanded" $A$ (i.e., $A \cup C$) should still be contained within the "expanded" $B$ (i.e., $B \cup C$). Any element that was originally in $A$ is now in $A \cup C$ and also in $B \cup C$. Any element that was in $C$ is now in $A \cup C$ and also in $B \cup C$. This suggests the statement is true.

To formally prove that a set $X$ is a subset of a set $Y$ (i.e., $X \subset Y$), the standard method is to show that for any arbitrary element $x$, if $x \in X$, then it must also be true that $x \in Y$. This is the element-wise proof technique.

Step-by-step Justification

We will assume the premise ($A \subset B$) and then logically deduce the conclusion ($A \cup C \subset B \cup C$).

1. Assume the premise: We are given that $A \subset B$.

   Important

   The definition of $A \subset B$ means that for any element $x$, if $x \in A$, then $x \in B$.

2. State the goal: We need to prove that $A \cup C \subset B \cup C$. To do this, we must show that every element in $A \cup C$ is also an element in $B \cup C$.

3. Consider an arbitrary element: Let $x$ be an arbitrary element such that $x \in A \cup C$.

4. Apply the definition of union: By the definition of the union of sets, if $x \in A \cup C$, it means that $x$ is in $A$ or $x$ is in $C$ (or both). We can write this as:

   $x \in A \quad \text{or} \quad x \in C$ …