Question

Let \(X\) be a topological space, \(Y\) a Hausdorff space, and \(f, g\) continuous maps from \(X\) to \(Y\). a. \(\\{x: f(x)=g(x)\\}\) is closed. b. If \(f=g\) on a dense subset of \(X\), then \(f=g\) on all of \(X\).

Short answer

(a) \(\{x : f(x) = g(x)\}\) is closed; (b) if \(f=g\) on a dense subset, then \(f=g\) on all of \(X\).

Step-by-step solution

Understanding the Problem

We have a topological space \(X\) and a Hausdorff space \(Y\), with continuous maps \(f\) and \(g\) from \(X\) to \(Y\). We need to show that the set \(\{x : f(x) = g(x)\}\) is closed, and that if \(f\) equals \(g\) on a dense subset of \(X\), then \(f = g\) on all of \(X\).

Show \(f(x) = g(x)\) is closed

To show that \(\{x : f(x) = g(x)\}\) is closed, note that for \(f(x) = g(x)\), we want \(f(x)\) and \(g(x)\) to converge to the same point in \(Y\). In a Hausdorff space, the set \( \{y : f(x) eq g(x)\} \), where points \(f(x)\) and \(g(x)\) differ, is an open set. This is because for any \( y_1 eq y_2 \) in \(Y\), there exist disjoint open sets containing them. Thus, the complement of our desired set, \( \{x : f(x) eq g(x)\} \), is open, which means \( \{x : f(x) = g(x)\} \) is closed.

Step 2: Show \(f=g\) on all of \(X\) if \(f=g\) on dense subset

Consider \(A\), the dense subset of \(X\) where \(f(x) = g(x)\). For any \(x \in X\), there's a sequence \((x_n)\) in \(A\) converging to \(x\) because \(A\) is dense. Since \(f\) and \(g\) are continuous, \(f(x_n) \to f(x)\) and \(g(x_n) \to g(x)\). But \(f(x_n) = g(x_n)\) for all \(n\), so \(f(x) = g(x)\). Thus, since \(f\) and \(g\) coincide on a set whose closure is all of \(X\), they coincide on all of \(X\).

Key concepts

Hausdorff Space

In topology, a Hausdorff space, also called a $T_2$ space, is a type of topological space where any two distinct points have disjoint neighborhoods. This means you can separate any two points with open sets. This property is useful in analyzing the behavior of continuous functions and constructing closed sets.

A practical implication of Hausdorff spaces is how they handle limits. In Hausdorff spaces, limits of sequences (or more general nets) converge to at most one point. This uniqueness simplifies many argumentations, like showing certain sets are closed. In the context of the given exercise, if two continuous functions from a topological space $X$ to a Hausdorff space $Y$ have the property of $(f(x) = g(x))$, it implies that we can define closed sets effectively, as seen in the next sections.

Continuous Functions

Continuous functions play a central role in topology and analysis, mapping elements from one topological space to another while preserving the concept of closeness. A function \(f: X \to Y\) is called continuous if for every open set \(V \subseteq Y\), the preimage \(f^{-1}(V)\) is an open set in \(X\).

The continuity ensures that the image under the function does not create gaps or jumps, maintaining the structure of space intact. This notion of continuity is pivotal when discussing when two mappings are identical over dense subsets as described in the exercise. Here, if \(f\) and \(g\) are continuous maps and \(f = g\) on a dense subset, continuous extensions ensure \(f = g\) on the whole domain \(X\).

Dense Subsets

A subset \(A\) of a topological space \(X\) is said to be dense in \(X\) if every point in \(X\) can be approximated arbitrarily closely by points from \(A\). Formally, the closure of \(A\), denoted by \(\overline{A}\), equals the entire space \(X\).

Dense subsets are crucial when discussing function equality across a space. If two continuous functions agree on a dense subset, their continuity ensures they will agree on the whole space. This is due to every point in \(X\) being a limit of points in the dense subset where the functions agree. Therefore, continuity forces agreement even outside the dense subset, as examined in the second part of the exercise solution.

Closed Sets

In topology, a closed set is one that contains all its limit points. Alternatively, the complement of a closed set is an open set. Closed sets are foundational in the study of topological spaces, offering a way to generalize the concept of closed intervals in real analysis.

In the provided exercise, we proved that the set \(\{x : f(x) = g(x)\}\) is closed. This was done by demonstrating that the set of points where \(f(x) eq g(x)\) is open, leveraging the Hausdorff property of the codomain and the concept of continuous functions. When an open set's complement is considered, it results in a closed set, underlining the importance of understanding the role closed sets play in topological arguments.