**
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.

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
\(X \times Y\).

Let's rephrase what this question is asking to make what we must prove a little clearer.