Category:Equivalence of Definitions of Continuous Mapping between Topological Spaces

This category contains pages concerning Equivalence of Definitions of Continuous Mapping between Topological Spaces:

Continuity at a Point

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 Neighborhood Inverse

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 $T_2$, $f^{-1} \sqbrk N$ is a neighborhood of $x$.

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 $\map f x$.

Continuity Everywhere

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 $f$.