Equivalence of Definitions of Continuous Mapping between Topological Spaces/Everywhere

== 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 $S_2$.
=== Definition by Pointwise Continuity ===
=== Definition by Open Sets ===
== Proof 1==
=== Sufficient Condition ===
Suppose that:
:$U_2 \in \tau_2 \implies f^{-1} \sqbrk {U_2} \in \tau_1$
Let $x \in S_1$.
Let $N \subseteq S_2$ be a neighborhood of $\map f x$.
By the definition of a neighborhood, there exists a $U_2 \in \tau_2$ such that $U_2 \subseteq N$, with $\map f x \in U_2$.
That is:
:$\exists U_2 \in \tau_2: \map f x \in U_2 \subseteq N \subseteq T_2$
We have:
:$\map f x \in U_2 \subseteq N \subseteq T_2 \implies f^{-1} \sqbrk {U_2} \in \tau_1$
Suppose further that $\map f x \in U_2$. 
By the definition of preimage, we have:
:$f^{-1} \sqbrk {U_2}$ is a neighborhood of $x$.
By Image of Subset under Mapping is Subset of Image, we have:
:$f \sqbrk {f^{-1} \sqbrk {U_2}} \subseteq U_2 \subseteq N$
Therefore, for every $x \in X$, we have that:
:For every neighborhood $N$ of $\map f x$ in $T_2$, there exists a neighborhood $f^{-1} \sqbrk {U_2}$ of $x$ in $T_1$ such that $f \sqbrk {f^{-1} \sqbrk {U_2}} \subseteq N$
Hence, for every $x \in X$, we have that:
: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 $f \sqbrk {U_1} \subseteq U_2$.
=== Necessary Condition ===
Now, suppose that $f$ is continuous at every point in $S_1$.
We wish to show that:
:$U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$
So, let $U \in \tau_2$.
Assume that $f^{-1} \sqbrk U$ is non-empty, otherwise $f^{-1} \sqbrk U = \O \in \tau_1$ by Empty Set is Element of Topology.
Let $x \in f^{-1} \sqbrk U$.
By the definition of continuity at a point, there exists a neighborhood $N$ of $x$ such that $f \sqbrk N \subseteq U$. 
By the definition of a neighborhood, there exists a $X \in \tau_1$ such that $x \in X \subseteq N$.
By Image of Subset under Mapping is Subset of Image:
:$f \sqbrk X \subseteq f \sqbrk N$
This gives:
:$f \sqbrk X \subseteq f \sqbrk N \subseteq U$
Let $\CC = \set {X \in \tau_1: f \sqbrk X \subseteq U}$.
Let $\ds H = \bigcup \CC$.
From the above argument:
:$f^{-1} \sqbrk U \subseteq H$
It follows directly from the definition of $H$ (and the definition of $\CC$) that:
:$H \subseteq f^{-1} \sqbrk U$
Hence:
:$H = f^{-1} \sqbrk U$.
By definition, $H$ is the union of open sets (of $S_1$). Hence $H$ is open by the definition of a topology.
== Proof 2 ==
=== Sufficient Condition ===
Suppose for every $x \in S_1$, the mapping $f$ is continuous at (the point) $x$ ( the topologies $\tau_1$ and $\tau_2$).
That is:
: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 $f \sqbrk {U_1} \subseteq U_2$.
Then:
:$U_2 \in \tau_2 \implies f^{-1} \sqbrk {U_2} \in \tau_1$
where $f^{-1} \sqbrk {U_2}$ denotes the preimage of $U_2$ under $f$.
Hence the result.
=== Necessary Condition ===
Suppose:
:$U_2 \in \tau_2 \implies f^{-1} \sqbrk {U_2} \in \tau_1$
where $f^{-1} \sqbrk {U_2}$ denotes the preimage of $U_2$ under $f$.
So, by definition of preimage:
:$U_2 \in \tau_2 \implies \set {x \in S_1: \exists t \in U_2: \map f x = t} \in \tau_1$
Then, for every $x \in S_1$, we have:
: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 $f \sqbrk {U_1} \subseteq U_2$.
Therefore, for every $x \in S_1$, the mapping $f$ is continuous at (the point) $x$ ( the topologies $\tau_1$ and $\tau_2$).
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 1$
Category:Equivalence of Definitions of Continuous Mapping between Topological Spaces