# Equivalence of Definitions of Continuous Mapping between Topological Spaces/Point

## Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.

Let $x \in S_1$.

The following definitions of the concept of **continuity at a point of a topological space** are equivalent:

### Definition using Open Sets

The mapping $f$ is **continuous at (the point) $x$** (with respect to the topologies $\tau_1$ and $\tau_2$) if and only if:

- For every open set $U_2$ of $T_2$ such that $\map f x \in U_2$, there exists an open set $U_1$ of $T_1$ such that $x \in U_1$ and $f \sqbrk {U_1} \subseteq U_2$.

### Definition using Neighborhoods

The mapping $f$ is **continuous at (the point) $x$** (with respect to the topologies $\tau_1$ and $\tau_2$) if and only if:

- For every neighborhood $N$ of $\map f x$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \sqbrk M \subseteq N$.

### Definition using Filters

The mapping $f$ is **continuous at (the point) $x$** if and only if:

- for any filter $\FF$ on $T_1$ that converges to $x$, the corresponding image filter $f \sqbrk \FF$ converges to $\map f x$.

## Proof

### $(1)$ implies $(2)$

Let $f$ be a continuous mapping defined using open sets.

Then by definition:

- For every open set $U_2$ of $T_2$ such that $\map f x \in U_2$, there exists an open set $U_1$ of $T_1$ such that $x \in U_1$ and $f \sqbrk {U_1} \subseteq U_2$.

Let $N$ be a neighborhood of $\map f x$ in $T_2$.

Then by definition of neighborhood, there exists an open set $U_2$ of $T_2$ such that $\map f x \in U_2$ and $U_2 \subseteq N$.

As $U_2$ is an open set $U_2$ of $T_2$, $f^{-1} \sqbrk {U_2}$ is an open set of $T_1$.

As $\map f x \in U_2$, it follows by definition of preimage of $U_2$ that $x \in f^{-1} \sqbrk {U_2}$.

Thus $f^{-1} \sqbrk {U_2}$ is a neighborhood of $x$ in $T_1$, which we relabel $M$.

From Image of Preimage under Mapping:

- $f \sqbrk M \subseteq U_2$

But we have that $U_2 \subseteq N$.

Thus we have demonstrated the existence of a neighborhood $M$ of $x$ such that $f \sqbrk M \subseteq N$

That is, $f$ is a continuous mapping defined using neighborhoods.

$\Box$

### $(2)$ implies $(1)$

Let $f$ be a continuous mapping defined using neighborhoods.

Then by definition:

- For every neighborhood $N$ of $\map f x$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \sqbrk M \subseteq N$.

Let $U_2$ be an open set of $T_2$ such that $\map f x \in U_2$.

By definition, $U_2$ is a neighborhood of $\map f x$.

Hence there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \sqbrk M \subseteq U_2$.

By definition, there exists an open set $U_1 \subseteq M$ such that $x \in U_1$.

That is, there exists an open set $U_1$ of $T_1$ such that $x \in U_1$ and $U_1 \subseteq M$.

By Image of Subset is Subset of Image:

- $f \sqbrk {U_1} \subseteq f \sqbrk M$

and so:

- $f \sqbrk {U_1} \subseteq U_2$

That is, $f$ is a continuous mapping defined using open sets.

$\Box$

### $(2)$ implies $(3)$

Let $f$ be a continuous mapping defined using neighborhoods.

Then by definition:

- For every neighborhood $N$ of $\map f x$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \sqbrk M \subseteq N$.

Let $\FF$ be a filter on $T_1$ that converges to $x$.

Let $N \subseteq S_2$ be a neighborhood of $\map f x$ in $T_2$.

Then $f^{-1} \sqbrk N$ is a neighborhood of $x$ in $T_1$.

Because $\FF$ converges to $x$, this implies that $f^{-1} \sqbrk N \in \FF$.

By the definition of $f \sqbrk \FF$ it follows that $N \in f \sqbrk \FF$.

Thus $f \sqbrk \FF$ converges to $\map f x$.

That is, $f$ is a continuous mapping defined using filters.

$\Box$

### $(3)$ implies $(2)$

Let $f$ be a continuous mapping defined using filters.

That is:

- for any filter $\FF$ on $T_1$ that converges to $x$, the corresponding image filter $f \sqbrk \FF$ converges to $\map f x$.

Let $\NN_x$ be the neighborhood filter of $x$:

- $\NN_x := \leftset {M \subseteq S_1: M}$ is a neighborhood of $\rightset x$

From Neighborhood Filter of Point is Filter, we have that $\NN_x$ is a filter on $S_1$.

Let $\FF := \NN_x$.

By the definition of convergent filter, $\FF$ converges to $x$.

By assumption this implies that $f \sqbrk \FF$ converges to $\map f x$.

Now let $N \subseteq S_2$ be a neighborhood of $\map f x$.

Then:

- $N \in f \sqbrk \FF$

By definition of image of subset under mapping:

- $f^{-1} \sqbrk N \in \FF = \NN_x$

Thus $f^{-1} \sqbrk N$ is a neighborhood of $x$.

Therefore $f$ is a continuous mapping defined using neighborhoods.

$\blacksquare$