## 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 $$f$$.

$\tag*{\blacksquare}$