Equivalence of Definitions of Continuous Mapping between Topological Spaces/Everywhere

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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 $S_2$.


The following definitions of the concept of everywhere continuous mapping between topological spaces are equivalent:

Definition by Pointwise Continuity

The mapping $f$ is continuous everywhere (or simply continuous) if and only if $f$ is continuous at every point $x \in S_1$.


Definition by Open Sets

The mapping $f$ is continuous on $S_1$ if and only if:

$U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$

where $f^{-1} \sqbrk U$ denotes the preimage of $U$ under $f$.


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

$\Box$


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.

$\blacksquare$


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Proof 2

Sufficient Condition

Suppose for every $x \in S_1$, the mapping $f$ is continuous at (the point) $x$ (with respect to 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 $f \sqbrk {U_1} \subseteq U_2$.


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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:

$U_2 \in \tau_2 \implies \set {x \in S_1: \exists t \in U_2: \map f x = t} \in \tau_1$



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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) $x$ (with respect to the topologies $\tau_1$ and $\tau_2$).

Hence the result.

Sources

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