Continuity Theorem in Metric Spaces

This theorem establishes the connection between the continuity of a function between metric spaces and the epsilon-delta definition.

A function $f$ mapping from one metric space $(X, d_X)$ to another $(Y, d_Y)$ is continuous if it satisfies the following conditions:

Take a point $x \in X$ and select a small positive value $\varepsilon > 0$, which represents how close the points in the image of $f$ should be.
There exists another positive value $\delta > 0$, representing how close we can get to $x$ in the space $X$.
If a point $x'$ is sufficiently close to $x$—that is, the distance between $x$ and $x'$, measured by $d_X$, is less than $\delta$: $$ d_X < \delta $$ then the points $f(x)$ and $f(x')$ will also be close in the space $Y$, with their distance $d_Y$ less than $\varepsilon$: $$ d_Y < \varepsilon $$

In simple terms, this formalizes the idea that a continuous function doesn't "jump" abruptly: small movements in the domain ($X$) result in small movements in the codomain ($Y$).

This is often referred to as the "epsilon-delta definition of continuity for metric spaces" or the "equivalence of continuity and the epsilon-delta property in metric spaces."

It’s essentially the same concept of continuity taught in Calculus 1, but generalized to the broader framework of metric spaces.

Note: The definition of continuity taught in Calculus 1, which deals with $\mathbb{R}$ or $\mathbb{R}^n$, is a specific case of this more general definition. In Calculus 1, a function $f : \mathbb{R} \to \mathbb{R}$ is continuous at a point $x \in \mathbb{R}$ if, for every $\varepsilon > 0$, there exists a $\delta > 0$ such that if $|x - x'| < \delta$, then $|f(x) - f(x')| < \varepsilon$. Here, the standard distances are used: $$ d_X(x, x') = |x - x'| $$ $$ d_Y(f(x), f(x')) = |f(x) - f(x')| $$ The definition for metric spaces applies to any function between metric spaces, not just functions on $\mathbb{R}$. However, the fundamental idea remains the same: "small changes in the input result in small changes in the output."

An Illustrative Example
The Proof

An Illustrative Example

Let’s consider two metric spaces:

Domain: $X = \mathbb{R}$, with the standard metric $d_X(x, x') = |x - x'|$.
Codomain: $Y = \mathbb{R}$, with the standard metric $d_Y(y, y') = |y - y'|$.

Let $f : \mathbb{R} \to \mathbb{R}$ be defined as:

$$ f(x) = 2x $$

We will verify that $f(x) = 2x$ is continuous using both the open set definition and the epsilon-delta definition, demonstrating their equivalence as stated in the theorem.

1] Continuity via Open Sets

In the topology induced by the standard metric, a set $V \subseteq Y$ is open if, for every $y \in V$, there exists $\varepsilon > 0$ such that the open ball $B_Y(y, \varepsilon) = \{y' \in Y \mid |y - y'| < \varepsilon\}$ is contained within $V$.

Let $V \subseteq Y$ be an open set. The preimage $f^{-1}(V)$ is defined as:

$$ f^{-1}(V) = \{x \in X \mid f(x) \in V\} $$

Since $f(x) = 2x$:

$$ f^{-1}(V) = \{x \in \mathbb{R} \mid 2x \in V\} $$

For every $y \in V$, there exists $\varepsilon > 0$ such that $B_Y(y, \varepsilon) \subseteq V$.

This implies that for $x \in f^{-1}(V)$, there exists $\delta = \varepsilon / 2$ such that the open ball $B_X(x, \delta)$ is entirely contained within $f^{-1}(V)$.

Therefore, the preimage of any open set in $Y$ is open in $X$, confirming that $f(x) = 2x$ is continuous under the topological definition.

2] Continuity via the Epsilon-Delta Definition

Let $x \in X$ and $\varepsilon > 0$. We need to find $\delta > 0$ such that if $|x - x'| < \delta$, then $|f(x) - f(x')| < \varepsilon$.

$$ f(x) = 2x \quad \text{and} \quad f(x') = 2x', \quad \text{thus:} $$

$$ |f(x) - f(x')| = |2x - 2x'| = 2|x - x'| $$

To ensure $|f(x) - f(x')| < \varepsilon$, we set:

$$ \delta = \frac{\varepsilon}{2} $$

If $|x - x'| < \delta = \varepsilon / 2$, then $|f(x) - f(x')| < \varepsilon$.

This confirms the continuity of $f$ using the epsilon-delta definition.

3] Conclusion

From this example, we conclude that:

The continuity of $f(x) = 2x$ ensures that the preimage of any open set is open.
The topological definition of continuity and the epsilon-delta definition are equivalent, as shown through this direct correspondence.

The Proof

We aim to establish the equivalence between two definitions of continuity for a function $f : X \to Y$, where $X$ and $Y$ are metric spaces:

Open set definition: $f$ is continuous if the preimage $f^{-1}(U)$ is open in $X$ for every open set $U \subseteq Y$.
Neighborhood definition: For every $x \in X$ and every open set $U \subseteq Y$ containing $f(x)$, there exists a neighborhood $V$ of $x$ in $X$ such that $f(V) \subseteq U$.

1] Proving the Open Set Definition Implies the Neighborhood Definition

Assume $f$ is continuous under the open set definition, meaning $f^{-1}(U)$ is open in $X$ for every open set $U \subseteq Y$.

Let $x \in X$ and $U \subseteq Y$ be an open set such that $f(x) \in U$.

Since $f^{-1}(U)$ is open in $X$, there exists a neighborhood $V$ of $x$ in $X$ such that $V \subseteq f^{-1}(U)$.

This implies $f(V) \subseteq U$, satisfying the neighborhood definition.

2] Proving the Neighborhood Definition Implies the Open Set Definition

Assume that for every $x \in X$ and every open set $U \subseteq Y$ containing $f(x)$, there exists a neighborhood $V$ of $x$ in $X$ such that $f(V) \subseteq U$.

To prove that $f^{-1}(W)$ is open in $X$ for any open set $W \subseteq Y$, consider a point $x \in f^{-1}(W)$. This means $f(x) \in W$.

Since $W$ is open and contains $f(x)$, by assumption, there exists a neighborhood $V$ of $x$ in $X$ such that $f(V) \subseteq W$.

Thus, $V \subseteq f^{-1}(W)$, proving $f^{-1}(W)$ is open in $X$.

We have shown that a function $f : X \to Y$ is continuous under the open set definition if and only if it satisfies the neighborhood definition.

And so forth.