Bounded Set A: Find Upper And Lower Bounds

Alright guys, let's dive into a fun little math problem! We need to show that the set $A = \{x \in \mathbb{R} : |x - 2| + |x - 1| < 5\}$ is bounded, and then pinpoint an upper bound (a majorante) and a lower bound (a minorante) for this set. Buckle up, it's gonna be a smooth ride!

Understanding the Set A

First off, let's get a grip on what the set A actually represents. It's a set of real numbers, $x$, that satisfy the inequality $|x - 2| + |x - 1| < 5$. The absolute value bits can seem a little daunting, but we'll break it down. Essentially, $|x - 2|$ is the distance between $x$ and 2, and $|x - 1|$ is the distance between $x$ and 1. The inequality is telling us that the sum of these distances must be less than 5.

Why is this important? Because this constraint on the sum of distances is what will allow us to show that the set is bounded. A set is bounded if it's contained within some finite interval, meaning there's a maximum and a minimum value beyond which no element of the set can exist. Let's explore different scenarios to remove those absolute values.

To handle the absolute values, we need to consider different intervals for $x$:

1. Case 1: $x < 1$

   If $x$ is less than 1, then both $x - 2$ and $x - 1$ are negative. So, $|x - 2| = -(x - 2) = 2 - x$ and $|x - 1| = -(x - 1) = 1 - x$. The inequality becomes:

   $(2 - x) + (1 - x) < 5$

   $3 - 2x < 5$

   $-2x < 2$

   $x > -1$

   So, in this case, we have $-1 < x < 1$.

2. Case 2: $1 \\\leq x < 2$

   If $x$ is between 1 and 2 (including 1 but not including 2), then $x - 2$ is negative and $x - 1$ is non-negative. So, $|x - 2| = -(x - 2) = 2 - x$ and $|x - 1| = x - 1$. The inequality becomes:

   $(2 - x) + (x - 1) < 5$

   $1 < 5$

   This is always true! So, all $x$ in the interval $[1, 2)$ satisfy the inequality.

3. Case 3: $x \\geq 2$

   If $x$ is greater than or equal to 2, then both $x - 2$ and $x - 1$ are non-negative. So, $|x - 2| = x - 2$ and $|x - 1| = x - 1$. The inequality becomes:

   $(x - 2) + (x - 1) < 5$

   $2x - 3 < 5$

   $2x < 8$

   $x < 4$

   So, in this case, we have $2 \\leq x < 4$.

Combining all these cases, we find that the set A is the interval $(-1, 4)$.

Proving the Set A is Bounded

Okay, now that we know A is the interval $(-1, 4)$, proving it's bounded becomes super straightforward. A set is bounded if it's both bounded above and bounded below. In other words, there exists a real number $M$ such that $x \\leq M$ for all $x$ in the set (bounded above), and there exists a real number $m$ such that $x \\geq m$ for all $x$ in the set (bounded below).

In our case, for the set $A = (-1, 4)$:

• Upper Bound: Any number greater than or equal to 4 is an upper bound. For example, 4, 4.5, 5, 100, etc., are all upper bounds. The least upper bound (also called the supremum) is 4.
• Lower Bound: Any number less than or equal to -1 is a lower bound. For example, -1, -1.5, -2, -100, etc., are all lower bounds. The greatest lower bound (also called the infimum) is -1.

Therefore, the set A is bounded.

Identifying a Majorante (Upper Bound) and a Minorante (Lower Bound)

Now, let's explicitly state an upper bound and a lower bound for the set A.

• Majorante (Upper Bound): We can choose 4 as an upper bound. Every element $x$ in A satisfies $x < 4$, so 4 is indeed an upper bound. Another valid choice would be 5 or any number greater than 4.
• Minorante (Lower Bound): We can choose -1 as a lower bound. Every element $x$ in A satisfies $x > -1$, so -1 is a lower bound. Another valid choice would be -2 or any number less than -1.

In summary:

• The set $A = \{x \in \mathbb{R} : |x - 2| + |x - 1| < 5\}$ is the interval $(-1, 4)$.
• The set A is bounded.
• An upper bound (majorante) for A is 4 (or any number greater than 4).
• A lower bound (minorante) for A is -1 (or any number less than -1).

Why Bounded Sets Matter

So, why all the fuss about bounded sets? Well, the concept of boundedness is fundamental in real analysis and many other areas of mathematics. Bounded sets have nice properties that unbounded sets don't. For example:

• Convergence: Bounded sequences (a sequence is just a special kind of set) are more likely to converge. The Bolzano-Weierstrass theorem guarantees that every bounded sequence in $\\mathbb{R}$ has a convergent subsequence.
• Completeness: The completeness axiom of the real numbers, which is crucial for many proofs, relies on the idea of bounded sets. It states that every non-empty set of real numbers that is bounded above has a least upper bound (supremum).
• Optimization: When you're trying to find the maximum or minimum value of a function, it often helps to restrict your attention to a bounded set. This can make the problem much easier to solve.

Alternative Approaches

While we solved this problem by breaking it down into cases based on the absolute values, there are other ways you could approach it. For instance, you could graph the function $f(x) = |x - 2| + |x - 1|$ and then find the values of $x$ for which $f(x) < 5$. The graph would be piecewise linear, and you could easily see the interval where the inequality holds.

Another approach involves using the triangle inequality, although it might not be the most direct route in this particular case. The triangle inequality states that $|a + b| \\leq |a| + |b|$ for any real numbers $a$ and $b$.

Conclusion

So there you have it! We've successfully shown that the set $A = \{x \in \mathbb{R} : |x - 2| + |x - 1| < 5\}$ is bounded, and we've identified a majorante (upper bound) and a minorante (lower bound) for the set. Remember, the key to tackling problems with absolute values is to break them down into cases where you can get rid of the absolute value signs. Keep practicing, and you'll become a pro at these types of problems in no time!

Understanding bounded sets is a cornerstone of real analysis and provides a solid foundation for more advanced topics. So, keep exploring, keep questioning, and keep learning! Math is an adventure, and there's always something new to discover. Keep your curiosity alive, guys!