Question

Show that a map \(f \cdot X \rightarrow Y\) between two topological spaces \(X\) and \(Y\) is contintous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for all sets \(U \subseteq X\) Show that \(f\) is a lwomeomorplism only if \(f(\bar{U})=\overline{f(U)}\) for all sets \(U \subseteq X\)

Short answer

So the function \(f: X \rightarrow Y\) is continuous if and only if for any subset \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). And \(f\) is a homeomorphism if and only if for any subset \(U \subseteq X\), \(f(\bar{U})= \overline{f(U)}\). Therefore, it's been demonstrated that these conditions are both necessary and sufficient for continuity and homeomorphism.

Step-by-step solution

Prove \(f\) is Continuous

Assume that for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Let \(V\) be an arbitrary closed subset of \(Y\). We must prove that \(f^{-1}(V)\) is also closed in \(X\). The complement \(X - f^{-1}(V)\) is an arbitrary subset of \(X\). Using the assumption, we have \(f(\overline{X - f^{-1}(V)}) \subseteq \overline{f(X - f^{-1}(V))}\). As \(f^{-1}(V)\) equals \(X - \overline{X - f^{-1}(V)}\), it follows that \(f^{-1}(V)\) is closed.

Show the Bi-conditional for the Continuity of \(f\)

Now we have to show the reverse, i.e., if \(f\) is continuous then for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Assume \(f\) is continuous, and let \(U\) be an arbitrary subset of \(X\). We must prove that \(f(\bar{U}) \subseteq \overline{f(U)}\). Since \(f\) is continuous, \(f(\bar{U})\) is closed in \(Y\), which means \(f(\bar{U}) \subseteq \overline{f(U)}\). Therefore, \(f\) is continuous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for every \(U \subseteq X\).

Prove \(f\) is a Homeomorphism

Again, let's assume that for every set \(U \subseteq X\), \(f(\bar{U}) = \overline{f(U)}\). From Step 2, it already follows that \(f\) is continuous. To prove that \(f\) is a homeomorphism, we must show that its inverse \(f^{-1}\) is also continuous. By applying the condition to \(f^{-1}\), we get that \(f^{-1}(\bar{V}) = \overline{f^{-1}(V)}\) for every \(V \subset Y\), thus proving that \(f^{-1}\) is continuous.

Show the Bi-conditional for Homeomorphism of \(f\)

For the reverse, assume that \(f\) is a homeomorphism. As a result, both \(f\) and \(f^{-1}\) are continuous (From Steps 1 and 3). Now we must prove that \(f(\bar{U}) = \overline{f(U)}\) for every \(U \subset X\). We already know from Step 2 that \(f(\bar{U}) \subseteq \overline{f(U)}\). For the reverse inclusion, denote \(V = f(U)\). As \(f^{-1}\) is continuous, we have \(f^{-1}(\overline{V}) \subseteq \overline{f^{-1}(V)} = \overline{U}\). As \(f(\overline{U}) \supseteq f(f^{-1}(\overline{V})) = \overline{V}\), we have proved the required.

Key concepts

Continuity

Continuity in topological spaces is a fundamental concept that extends the notion of continuity of real-valued functions familiar from calculus. In the context of topology, a function \( f: X \to Y \) between two topological spaces \( X \) and \( Y \) is called continuous if the pre-image of every open set in \( Y \) is also open in \( X \). This means that small changes in \( X \) result in small changes in \( Y \), maintaining the structure of the topological spaces.

• To prove continuity, one way is to show that the pre-image of a closed set is also closed. This approach is particularly useful in showing that \( f^{-1}(V) \) is closed in \( X \) if \( V \) is closed in \( Y \).
• A key result used in proving continuity is that if \( f(\bar{U}) \subseteq \overline{f(U)} \) for every subset \( U \subseteq X \), then \( f \) is continuous. This ensures that images of "limit points" are still limit points, preserving closure properties under the function.

In topological spaces, continuity ensures a kind of smooth transition between elements of the spaces under the function \( f \).

Homeomorphism

A homeomorphism is a special type of function in topology that describes a very strong form of equivalence between topological spaces. Essentially, spaces that are homeomorphic are considered the same in terms of their topological properties, meaning they can be continuously transformed into each other without tearing or gluing.For a function \( f: X \to Y \) to be a homeomorphism, it must satisfy two key conditions:

• \( f \) itself must be continuous, preserving the open set structure from \( X \) to \( Y \).
• The inverse function \( f^{-1}: Y \to X \) must also be continuous, ensuring the structure is preserved when mapping back from \( Y \) to \( X \).

Additionally, bijection is required to ensure that each element of \( Y \) corresponds uniquely to an element of \( X \) and vice versa.Through homeomorphisms, topologists identify spaces that, while they may appear different geometrically, share the same fundamental structure.

Closed Sets

In the realm of topology, closed sets are a basic building block, alongside open sets, crucial for understanding more complex structures.A set is closed if it contains all its limit points. This can be expressed as \( X - U \) is open when \( U \) is closed, where \( X \) is the universal set containing \( U \). Closed sets, therefore, are complements of open sets and contribute to defining continuity in functions.Closed sets have significant importance:

• They allow us to define the closure of a set \( U \), denoted \( \bar{U} \), which is the smallest closed set containing \( U \).
• Understanding how functions operate on closed sets helps in discussing properties like continuity and homeomorphism, particularly seen in \( f(\bar{U}) \subseteq \overline{f(U)} \).

The closed set concept is pivotal in the intuition of how spaces relate and transform under topological functions.

Continuous Maps

A continuous map between two topological spaces ensures a seamless transition of topological structure from one space to another. This is not just about preserving distances as in metric spaces but more about maintaining the fundamental shape or structure of a set.When a function \( f: X \to Y \) is continuous, any open (or closed) set in \( X \) is transformed into an open (or closed) set in \( Y \) in such a way that nearby points remain nearby after the mapping.Continuous maps are characterized by:

• Being able to take pre-images of open sets to open sets (or equivalently, closed to closed, which simplifies the investigation of continuity in closed forms).
• Ensuring that "nearness" or "closeness" in one space translates into "nearness" in the image space. This is demonstrated by \( f(\bar{U}) \subseteq \overline{f(U)} \).

Understanding continuous maps is crucial for studying how different topological spaces relate to each other and exploring more complex concepts like homeomorphisms.