Open Disconnected Sets Under Continuous 1-1 Functions

Hey guys! Let's dive into a fascinating problem from real analysis. We're going to explore how one-to-one continuous functions behave when they map open and disconnected sets. This is a classic topic that really tests our understanding of continuity, openness, and connectedness in the context of functions between Euclidean spaces. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into the nitty-gritty, let's make sure we're all on the same page with the problem statement. We're given a function f that maps from $\mathbb{R}^n$ to $\mathbb{R}^m$. A few key properties define this function:

• One-to-one: This means that each element in the domain maps to a unique element in the codomain. In simpler terms, no two different points in $\mathbb{R}^n$ get mapped to the same point in $\mathbb{R}^m$.
• Continuous: This is a big one! Continuity means that small changes in the input result in small changes in the output. More formally, for any point x in $\mathbb{R}^n$ and any neighborhood around f(x) in $\mathbb{R}^m$, we can find a neighborhood around x in $\mathbb{R}^n$ such that the image of this neighborhood under f is contained in the neighborhood around f(x).

We're also given a set A that lives in $\mathbb{R}^n$, and A has two important properties:

• Open: An open set contains a neighborhood around each of its points. Think of it like this: if you pick any point in an open set, you can always draw a little ball around that point that's entirely contained within the set.
• Disconnected: A disconnected set can be expressed as the union of two non-empty, disjoint open sets. In other words, you can split the set into two pieces that are separated from each other.

Our mission, should we choose to accept it, is to prove that the image of A under f, denoted as f(A), is also open and disconnected, but this time within the space f($\mathbb{R}^n$). This means we need to show two things:

1. f(A) is open in f($\mathbb{R}^n$).
2. f(A) is disconnected.

This problem beautifully combines concepts from topology and real analysis. It's a great exercise in thinking about how continuity preserves certain properties of sets, and how one-to-one functions ensure that the structure of the domain is reflected in the range. Let's break down how we can tackle this proof step by step.

Proving $f(A)$ is Open

Okay, let's start by showing that f(A) is open in f($\mathbb{R}^n$). This is a crucial step because openness is a fundamental topological property. Remember, a set is open if every point in the set has a neighborhood entirely contained within the set.

To prove f(A) is open in f($\mathbb{R}^n$), we need to show that for any point y in f(A), there exists an open set V in f($\mathbb{R}^n$) such that y belongs to V and V is entirely contained in f(A). This might sound a bit abstract, but let's break it down.

Since y is in f(A), it means there exists a point x in A such that f(x) = y. Now, we know that A is an open set in $\mathbb{R}^n$. This means there exists an open ball (or an open n-dimensional sphere) centered at x, let's call it B(x, r), with radius r > 0, such that B(x, r) is entirely contained in A. Mathematically, we can write this as:

$B(x, r) = \{x' \in \mathbb{R}^n : ||x' - x|| < r\} \subseteq A$

Now, here's where the one-to-one and continuous nature of f comes into play. Since f is continuous, it's tempting to think that the image of an open ball is always open. However, this isn't always true in general. But because f is one-to-one and continuous on $\mathbb{R}^n$, a powerful result comes to our rescue: the Invariance of Domain Theorem. This theorem, in essence, tells us that a continuous, injective (one-to-one) map from an open subset of $\mathbb{R}^n$ into $\mathbb{R}^n$ maps open sets to open sets. While the general Invariance of Domain Theorem can be quite deep, for our purposes, it provides the crucial link we need.

Applying the Invariance of Domain Theorem to our scenario, we can conclude that f(B(x, r)) is an open set in $\mathbb{R}^m$. But we need to show that f(A) is open in f($\mathbb{R}^n$), not necessarily in $\mathbb{R}^m$ itself. To bridge this gap, we consider the intersection of f(B(x, r)) with f($\mathbb{R}^n$). Let's define this intersection as:

$V = f(B(x, r)) \cap f(\mathbb{R}^n)$

This set V is open in f($\mathbb{R}^n$) because it's the intersection of an open set f(B(x, r)) in $\mathbb{R}^m$ with f($\mathbb{R}^n$), which inherits the subspace topology from $\mathbb{R}^m$.

Now, we need to show that V satisfies our criteria: it contains y and is contained in f(A). Since x belongs to B(x, r), f(x) = y belongs to f(B(x, r)). Also, since y is in f(A), it must be in f($\mathbb{R}^n$). Therefore, y is in the intersection V.

Finally, since B(x, r) is a subset of A, the image f(B(x, r)) is a subset of f(A). This means that V, being a subset of f(B(x, r)), is also a subset of f(A). Thus, we have found an open set V in f($\mathbb{R}^n$) that contains y and is contained in f(A). This completes the proof that f(A) is open in f($\mathbb{R}^n$).

Proving $f(A)$ is Disconnected

Great job, guys! We've successfully shown that f(A) is open. Now, let's tackle the second part of our mission: proving that f(A) is disconnected. This is where we'll leverage the fact that A itself is disconnected.

Recall that a set is disconnected if it can be written as the union of two non-empty, disjoint open sets. Since A is disconnected, there exist two open sets, let's call them U and V, such that:

1. A = U ∪ V
2. U ≠ ∅ and V ≠ ∅
3. U ∩ V = ∅

Now, our goal is to show that f(A) can also be expressed as the union of two non-empty, disjoint open sets within f($\mathbb{R}^n$). This will directly demonstrate that f(A) is indeed disconnected.

Let's consider the images of U and V under f. We define:

• f(U) = { f(x) | x ∈ U }
• f(V) = { f(x) | x ∈ V }

Since A = U ∪ V, and f is a function, it follows that:

$f(A) = f(U \cup V) = f(U) \cup f(V)$

This is a good start! We've expressed f(A) as the union of two sets, f(U) and f(V). Now, we need to show that these sets are open in f($\mathbb{R}^n$), non-empty, and disjoint.

Showing $f(U)$ and $f(V)$ are Open in $f(\mathbb{R}^n)$

This part is similar to what we did earlier when proving f(A) is open. Since U and V are open sets in $\mathbb{R}^n$, we can apply the same logic, utilizing the Invariance of Domain Theorem. For any point y in f(U), there exists a point x in U such that f(x) = y. Because U is open, there's an open ball B(x, r) contained in U. Applying the Invariance of Domain Theorem, f(B(x, r)) is open in $\mathbb{R}^m$. Therefore, the intersection f(B(x, r)) ∩ f($\mathbb{R}^n$) is open in f($\mathbb{R}^n$) and contains y. This shows that f(U) is open in f($\mathbb{R}^n$). The same argument applies to f(V), so it's also open in f($\mathbb{R}^n$).

Showing $f(U)$ and $f(V)$ are Non-Empty

This is straightforward. Since U and V are non-empty by definition (they're part of the disconnectedness condition for A), there exist points in each set. Because f is a function, the images of these points will be in f(U) and f(V), respectively. Therefore, both f(U) and f(V) are non-empty.

Showing $f(U)$ and $f(V)$ are Disjoint

This is where the one-to-one property of f shines. We need to show that f(U) ∩ f(V) = ∅. We'll use a proof by contradiction.

Suppose, for the sake of contradiction, that f(U) ∩ f(V) ≠ ∅. This means there exists a point y that belongs to both f(U) and f(V). Therefore, there exists a point x₁ in U such that f(x₁) = y, and there exists a point x₂ in V such that f(x₂) = y. But this implies that f(x₁) = f(x₂), where x₁ is in U and x₂ is in V. Since U and V are disjoint, x₁ ≠ x₂. However, this contradicts the fact that f is one-to-one! A one-to-one function cannot map two distinct points to the same point.

Therefore, our assumption that f(U) ∩ f(V) ≠ ∅ must be false. Hence, f(U) ∩ f(V) = ∅, meaning f(U) and f(V) are disjoint.

Conclusion for Disconnectedness

We've successfully shown that f(A) = f(U) ∪ f(V), where f(U) and f(V) are non-empty, disjoint, and open in f($\mathbb{R}^n$). This exactly fits the definition of a disconnected set. Therefore, f(A) is disconnected.

Final Thoughts

Wow, guys, we made it! We've proven that if f is a one-to-one continuous function from $\mathbb{R}^n$ to $\mathbb{R}^m$, and A is an open and disconnected subset of $\mathbb{R}^n$, then f(A) is open and disconnected in f($\mathbb{R}^n$). This was a fantastic journey through the concepts of continuity, openness, disconnectedness, and the power of the Invariance of Domain Theorem.

This problem highlights how continuous functions, especially those with the added property of being one-to-one, preserve topological properties. It’s a great example of how abstract mathematical concepts can come together to create elegant and insightful results. Keep exploring, keep questioning, and keep enjoying the beauty of mathematics!