Set Subtraction And Translation: Does Order Matter?

Hey guys! Today, we're diving into an interesting question in the realm of geometry, linear transformations, convex analysis, and sumsets. Specifically, we're going to explore the interplay between set subtraction and translation in $\mathbb{R}^n$. It's a fascinating topic that touches upon fundamental concepts in these areas.

Understanding Set Subtraction and Translation

Before we get to the core question, let's make sure we're all on the same page regarding the definitions of set subtraction and translation. This is crucial for understanding the nuances of the problem and avoiding any confusion later on. We will break down each operation separately and provide some simple examples to solidify the concepts. Think of this as laying the groundwork for our exploration.

Set Subtraction: What Does it Really Mean?

In the context of sets in $\mathbb{R}^n$, set subtraction (also known as the Minkowski difference) isn't quite the same as subtracting numbers. Instead, it involves taking the difference between all possible pairs of points, one from each set. Formally, if we have two sets $A$ and $B$ in $\mathbb{R}^n$, their set difference, denoted as $A - B$, is defined as:

$A - B = \{a - b \mid a \in A, b \in B\}$

In simpler terms, we take every point in set $A$ and subtract every point in set $B$ from it. The resulting set $A - B$ contains all these difference vectors. Let's look at a simple example in $\mathbb{R}^2$ to illustrate this. Suppose $A = \{(1, 2), (3, 4)\}$ and $B = \{(0, 1), (2, 0)\}$. Then, to find $A - B$, we perform the following subtractions:

• (1, 2) - (0, 1) = (1, 1)
• (1, 2) - (2, 0) = (-1, 2)
• (3, 4) - (0, 1) = (3, 3)
• (3, 4) - (2, 0) = (1, 4)

Therefore, $A - B = \{(1, 1), (-1, 2), (3, 3), (1, 4)\}$. Notice that the set difference can result in a set with more elements than either of the original sets, as we are considering all possible combinations of subtractions. This is a key characteristic of set subtraction that distinguishes it from element-wise subtraction.

Translation: Shifting Sets in Space

Translation, on the other hand, is a more intuitive operation. It simply involves shifting a set by a fixed vector. If we have a set $A$ in $\mathbb{R}^n$ and a vector $c \in \mathbb{R}^n$, the translation of $A$ by $c$, denoted as $A + c$, is defined as:

$A + c = \{a + c \mid a \in A\}$

In other words, we add the vector $c$ to every point in the set $A$. Geometrically, this corresponds to sliding the set $A$ along the direction of $c$ without changing its shape or orientation. Think of it as picking up the set and moving it to a new location.

Let's revisit our previous example and translate the set $A = \{(1, 2), (3, 4)\}$ by the vector $c = (1, -1)$. Then, $A + c$ is obtained by adding $(1, -1)$ to each point in $A$:

• (1, 2) + (1, -1) = (2, 1)
• (3, 4) + (1, -1) = (4, 3)

So, $A + c = \{(2, 1), (4, 3)\}$. As you can see, the translated set has the same number of elements as the original set, and its overall shape remains unchanged. The key takeaway here is that translation is a rigid motion, preserving the geometric properties of the set.

The Core Question: Can We Swap Subtraction and Translation?

Now that we have a solid understanding of set subtraction and translation, we can tackle the main question: Given sets $A, B \subset \mathbb{R}^n$ and a point $c \in \mathbb{R}^n$, is it always true that $(A + c) - B = A - (B + c)$? This question delves into the algebraic properties of these operations and how they interact with each other. It's not immediately obvious whether the order matters, so we need to investigate carefully.

Breaking Down the Equation

To answer this question, let's first break down what each side of the equation represents. On the left-hand side, $(A + c) - B$, we first translate the set $A$ by the vector $c$, resulting in the set $A + c$. Then, we perform set subtraction between $A + c$ and $B$. This means we take every point in $A + c$ and subtract every point in $B$ from it.

On the right-hand side, $A - (B + c)$, we first translate the set $B$ by the vector $c$, resulting in the set $B + c$. Then, we perform set subtraction between $A$ and $B + c$. This means we take every point in $A$ and subtract every point in $B + c$ from it.

The question, therefore, is whether these two processes always lead to the same result. Intuitively, one might think that the order shouldn't matter since we're essentially dealing with vector addition and subtraction, which are commutative and associative. However, the set operations introduce a level of complexity that requires a more rigorous analysis. It's crucial to remember that set subtraction is not simply element-wise subtraction; it involves considering all possible pairs of points.

A Detailed Proof

To determine whether the equation holds, we need to prove or disprove it. Let's start by attempting to prove it. We want to show that $(A + c) - B = A - (B + c)$. To do this, we'll show that each side is a subset of the other. This is a standard technique for proving set equality: if $X \subseteq Y$ and $Y \subseteq X$, then $X = Y$.

Proving $(A + c) - B \subseteq A - (B + c)$

Let $x \in (A + c) - B$. This means that there exists a point $a' \in A + c$ and a point $b \in B$ such that $x = a' - b$. Since $a' \in A + c$, there exists a point $a \in A$ such that $a' = a + c$. Substituting this into our equation for $x$, we get:

$x = (a + c) - b = a + c - b$

Now, we want to show that $x \in A - (B + c)$. This means we need to find a point $b' \in B + c$ such that $x = a - b'$. Since $b' \in B + c$, there exists a point $b \in B$ such that $b' = b + c$. Let's try to manipulate our equation for x to match this form.

We have $x = a + c - b$. We can rewrite this as:

$x = a - (b - c)$

However, this isn't quite in the form we need. We need to subtract something of the form $b + c$, not $b - c$. This is a potential roadblock in our proof.

Let's try a different approach. Instead of manipulating the equation directly, let's think about what it means for $x$ to be in $A - (B + c)$. It means there exists $a \in A$ and $b' \in B + c$ such that $x = a - b'$. Since $b' \in B + c$, there exists $b \in B$ such that $b' = b + c$. Therefore, $x = a - (b + c) = a - b - c$. This looks promising.

Comparing this to our previous expression for $x$, which was $x = a + c - b$, we see that they are not the same. This suggests that the inclusion $(A + c) - B \subseteq A - (B + c)$ might not hold in general.

Proving $A - (B + c) \subseteq (A + c) - B$

Let's try to prove the reverse inclusion. Let $x \in A - (B + c)$. This means there exists a point $a \in A$ and a point $b' \in B + c$ such that $x = a - b'$. Since $b' \in B + c$, there exists a point $b \in B$ such that $b' = b + c$. Substituting this into our equation for $x$, we get:

$x = a - (b + c) = a - b - c$

Now, we want to show that $x \in (A + c) - B$. This means we need to find a point $a' \in A + c$ and a point $b \in B$ such that $x = a' - b$. Since $a' \in A + c$, there exists a point $a \in A$ such that $a' = a + c$. Therefore, we want to see if we can write $x$ in the form $(a + c) - b$ for some $b \in B$.

We have $x = a - b - c$. We can rewrite this as:

$x = (a - c) - b$

However, this is not in the form we want. We need $a + c$, not $a - c$. Again, we encounter a roadblock. This further reinforces the idea that the equation might not hold in general.

Finding a Counterexample: Where the Equation Fails

Since our attempts to prove the equation have been unsuccessful, let's try to find a counterexample. A counterexample is a specific instance where the equation doesn't hold. This will definitively disprove the claim that the equation is always true.

Let's consider a simple example in $\mathbb{R}^1$. Let $A = \{0, 1\}$, $B = \{0\}$, and $c = 1$. Now, let's compute both sides of the equation:

• $A + c = \{0 + 1, 1 + 1\} = \{1, 2\}$
• $(A + c) - B = \{1 - 0, 2 - 0\} = \{1, 2\}$
• $B + c = \{0 + 1\} = \{1\}$
• $A - (B + c) = \{0 - 1, 1 - 1\} = \{-1, 0\}$

In this case, $(A + c) - B = \{1, 2\}$ and $A - (B + c) = \{-1, 0\}$. These sets are clearly not equal. Therefore, we have found a counterexample, which disproves the general equation! This is a significant finding.

Why Does the Equation Fail? The Role of Set Subtraction

So, why does the equation fail? The key lies in the nature of set subtraction. As we discussed earlier, set subtraction involves considering all possible differences between points in the two sets. This operation is not as straightforward as element-wise subtraction, and it doesn't always play nicely with translation.

In our counterexample, the translation by $c$ effectively shifts the set $A$, changing the possible differences we can obtain when subtracting $B$. Similarly, translating $B$ changes the points we are subtracting, which also affects the resulting set. The order in which we perform these operations matters because set subtraction is not a linear operation in the same way that simple addition and subtraction are.

Implications and Further Exploration

This result has important implications for working with sets and linear transformations. It highlights the fact that set operations, while seemingly simple, can have subtle and sometimes counterintuitive properties. It's a reminder that we need to be careful when manipulating sets and avoid making assumptions based solely on our intuition from working with numbers.

This exploration also opens up further questions. For example, under what conditions does the equation $(A + c) - B = A - (B + c)$ hold? Are there specific types of sets or vectors $c$ for which the order of subtraction and translation doesn't matter? These are interesting avenues for further investigation.

Conclusion: Order Matters!

In conclusion, we've shown that, in general, $(A + c) - B \neq A - (B + c)$ for sets $A, B \subset \mathbb{R}^n$ and a point $c \in \mathbb{R}^n$. We arrived at this conclusion by attempting to prove the equation, encountering roadblocks, and ultimately finding a counterexample. This highlights the importance of rigorous reasoning and the potential pitfalls of relying solely on intuition when dealing with set operations. Remember guys, in the world of set subtraction and translation, order matters! This understanding is crucial for anyone working in geometry, linear transformations, convex analysis, or any field that involves manipulating sets in Euclidean space.