Problem 6 - Analysis Now

Question

Let \(X, Y\), and \(Z\) be topological spaces and give \(X \times Y\) the product topology. Show that if a function \(f: X \times Y \rightarrow Z\) is continuous, then it is separately continuous in each variable [i.e., for each \(x\) in \(X\) the function \(y \rightarrow f(x, y)\) is continuous from \(Y\), to \(Z\), and similarly for each \(y\) in \(Y\) ]. Show by an example that the converse does not hold. Hint: Try \(f(x, y)=x y\left(x^{2}+y^{2}\right)^{-1 / 2}\) if \((x, y) \neq(0,0)\) and \(f(0,0)=0\).

Step-by-Step Solution

Answer

The function is separately continuous but not continuous in the product topology context.

Step 1: Continuity Definition Recap

First, remember that a function \( f: X \times Y \rightarrow Z \) is continuous if for any open set \( V \subseteq Z \), the preimage \( f^{-1}(V) \) is open in \( X \times Y \). In the product topology, an open set in \( X \times Y \) is a union of basis elements \( U \times V \), where \( U \) is open in \( X \) and \( V \) is open in \( Y \).

Step 2: Prove Separate Continuity with Continuity Assumption

Given a continuous function \( f: X \times Y \rightarrow Z \), fix a point \( x_0 \in X \). Consider the function \( g: Y \rightarrow Z \) given by \( g(y) = f(x_0, y) \). To prove \( g \) is continuous, take any open \( V \subseteq Z \) and consider \( g^{-1}(V) = \{ y \in Y \,|\, f(x_0, y) \in V \} \). This set must be open in \( Y \) because the preimage \( f^{-1}(V) \), which includes the set \( \{ (x, y) \in X \times Y \ | \ f(x, y) \in V \} \), is open in the product space, implying the section at \( x_0 \) is open. Similar logic shows continuity in the \( x \) coordinate for fixed \( y \).

Step 3: Choose an Example for Separate Not Joint Continuity

Consider the function \( f(x, y)=x y \left(x^{2}+y^{2}\right)^{-1 / 2} \) if \( (x, y) eq (0,0) \) and \( f(0,0)=0 \). This function seems naturally defined, but we need to check its continuity properties.

Step 4: Show the Example is Separately Continuous

Fix \( y = y_0 \) and analyze \( g: X \rightarrow Z \) defined by \( g(x) = f(x, y_0) \). For \( (x, y) eq (0, 0) \), continuity in \( x \) (holding \( y = y_0 \) fixed) shows continuity as cancelling terms \( x \) simplify the expression. A similar argument holds for continuity in \( y \) (holding \( x = x_0 \) fixed).

Step 5: Explain Non-General Continuity

To show \( f \) is not continuous at \( (0,0) \) in the product sense, observe that approaching \( (0,0) \) along lines where \( x = y \), \( f(x, y) \) approaches different values, disrupting the single limit requirement for continuity in a full sense.

Key Concepts

Product Topology

In topology, the concept of a "product topology" is crucial when dealing with Cartesian product spaces, such as \(X \times Y\). Given topological spaces \(X\) and \(Y\), the product topology on \(X \times Y\) is formed using these individual topologies. The product topology considers open sets to be unions of products of open sets from each space.
A basis for the product topology consists of sets which are Cartesian products \(U \times V\), where \(U\) is open in \(X\) and \(V\) is open in \(Y\). This means any open set in \(X \times Y\) can be expressed as a union of such basis sets.
This topology is very useful because it allows us to study continuity of functions from product spaces in a systematic way. By understanding how open sets are formed within the product space, one can deduce how functions behave across both dimensions of the product.

Separate Continuity

Separate continuity refers to a function \(f: X \times Y \rightarrow Z\) being continuous with respect to each variable separately. This means if you fix one variable, the function is still continuous in terms of the other variable.
For example, if you fix \(x = x_0\), then the function \(g(y) = f(x_0, y)\) should be continuous considering the variable \(y\) only. Similarly, if you fix \(y = y_0\), the continuity then needs to be checked for \(h(x) = f(x, y_0)\).
This concept is different from joint continuity, where we expect the function to behave continuously with respect to changes in both variables taken together.

Continuous Function

The notion of a continuous function is central in topology. A function \(f: X \rightarrow Z\) is called continuous if the preimage of every open set in \(Z\) is open in \(X\).
When considering functions from a product space \(X \times Y\) to \(Z\), continuity demands that for any open set \(V\) in \(Z\), the set \(f^{-1}(V)\) is open in the product topology of \(X \times Y\). This requirement ensures that as we take points close to each other in \(X \times Y\), their images under \(f\) remain close in \(Z\), preserving the structure of open sets through the function.

Preimage of Open Sets

In topology, when speaking about continuity, the concept of a preimage of open sets is foundational. Suppose you have a function \(f: X \rightarrow Y\) and an open set \(V\) in \(Y\). The preimage, \(f^{-1}(V)\), is the collection of all points in \(X\) that are mapped to \(V\) by \(f\).
This preimage must be open for \(f\) to be considered continuous. Intuitively, this means that neighborhoods (open sets) in the range should correspond to neighborhoods in the domain.
When applied to functions from product spaces, understanding preimages becomes more complex but remains guided by this same principle—ensuring open sets map back to open sets is what preserves continuity across spaces.