Continuous Functions
July 2020
Problem 1: Prove that for functions \(f: \mathbb{R} \rightarrow \mathbb{R}\), the \(\epsilon-\delta\) definition of continuity implies the open set definition.
The \(\epsilon-\delta\) definition: A map \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous if for all \(x\) in the domain,
\[\forall \epsilon > 0, \exists \ \delta > 0 \textrm{ such that } |y-x| < \delta \implies |f(y) - f(x)| < \epsilon\]
The open set definition: Let \(X\) and \(Y\) be topological spaces.
\(f: X \rightarrow Y\) is continuous if for each open subset
\(V\) of \(Y\), \(f^{-1}(V)\) is an open subset of \(X\).
Solution: For some \(x_0 \in \mathbb{R}\) and some \(\epsilon > 0\), let \(V = (f(x_0) - \epsilon, f(x_0) + \epsilon)\). We show that \(f^{-1}(V)\) is open. By the \(\epsilon-\delta\) definition, for any \(x \in f^{-1}(V)\), there exists a \(\delta > 0\) such that for any \(y \in (x - \delta, x + \delta)\), \(f(y) \in V\). This means that \((x - \delta, x + \delta)\) must be in \(f^{-1}(V)\). Since there is such an open set for every \(x \in f^{-1}(V)\), \(f^{-1}(V)\) is a union of open sets and is therefore open. Since any open set in the codomain is going to be the union of intervals like \(V\), this applies to any open set in the codomain and its inverse.
\[\tag*{$\blacksquare$}\]
Problem 11: Let \(F: X \times Y \rightarrow Z\). \(F\) is continuous in each variable separately if for each \(y_0\) in \(Y\), the map \(h: X \rightarrow Z\) defined by \(h(x)=F(x \times y_0)\) is continuous, and for each \(x_0\) in \(X\), the map \(k: Y \rightarrow Z\) defined by \(k(y) = f(x_0 \times y)\) is continuous. Show that if \(F\) is continuous, then \(F\) is continuous in each variable separately.
To prove this we can make use of Theorem 18.2 in the book, which states that if \(f:X \rightarrow Y\) is continuous and \(A\) is a subspace of \(X\), then the restricted function \(f|A: A \rightarrow Y\) is continuous. In this case, we are restricting the domain to either \(X \times {y_0}\), or \(x_0 \times Y\) where these new domains inherit the subspace topology from \(X \times Y\).
\[\tag*{$\blacksquare$}\]
Problem 13: Let \(A \subset X\); let \(f: A \rightarrow Y\) be continuous; let \(Y\) be Hausdorff. Show that if \(f\) may be extended to a continuous function \(g: \overline{A} \rightarrow Y\), then \(g\) is uniquely determined by \(f\).
Suppose there are two distinct functions \(g_1, g_2: \overline{A} \rightarrow Y\) such that \(g_i | A = f\). Since they are distinct, there must exist some \(x \in \overline{A}\) such that \(g_1(x) \neq g_2(x)\). Let \(U, V\) be disjoint open neighbourhoods in \(Y\) of \(g_1(x), g_2(x)\), respectively. We can define such neighbourhoods since \(Y\) is Hausdorff.
Since \(g_1, g_2\) are continuous,
\[S = g_1^{-1}(U), \ \ \ T = g_2^{-1}(V)\]
are both open neighbourhoods of \(x\) in \(\overline{A}\) (meaning \(S \cap T\) isn't empty). Note that because \(x \in \overline{A}\), all open neighbourhoods of \(x\) must contain points of \(A\).
Since \( \ g_1 | A = g_2 | A = f\),
\[ g_1 (S \cap A \cap T) = g_2 (S \cap A \cap T)\]
But this is a contradiction, since \(g_1(S) \cap g_2(T) \subset U \cap V = \varnothing\). Then there can't exist disjoint neighbourhoods of \(g_1(x), g_2(x)\) for any \(x\), which means \(g_1 = g_2\), and so \(g\) is uniquely determined by \(f\).
\[\tag*{$\blacksquare$}\]