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 $$X$$.

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.

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 $$X \times Y$$.
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 $$f$$.
Let's rephrase what this question is asking to make what we must prove a little clearer.