Continuous Maps on Topological Spaces

Continuous Mappings on Topological Spaces

Recall from the Local Bases of a Point in a Topological Space page that if $(X, \tau)$ is a topological space then a local basis of a point $x \in X$ is a collection $\mathcal B_x$ of open neighbourhoods of $x$ such that for each $U \in \tau$ with $x \in U$ there exists a $B \in \mathcal B_x$ such that $B$ is contained in $U$, that is:

(1)

\begin{align} \quad x \in B \subseteq U \end{align}

Definition: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a mapping from $X$ to $Y$. A mapping $f$ is said to be Continuous at the point $a \in X$ if there exists local bases of $a$ and $f(a)$, denote them $\mathcal B_a$ and $\mathcal B_{f(a)}$ such that for all $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_a$ such that $f(B') \subseteq B$.

In other words, $f$ is continuous at $a \in X$ if there exists local bases of $\mathcal B_a$ of $a$ and $\mathcal B_{f(a)}$ of $f(a)$ such that for every set set $B \in \mathcal B_{f(a)}$ there exists a $B' \in \mathcal B_a$ whose image, $f(B')$ is contained in $B$, i.e., $f(B') \subseteq B$.

Note that $f(a)$ denotes an element in $Y$, i.e., $f(a) \in Y$, while $f(B')$ denotes the image of the set $B'$ under $f$.

We will later see that there are many equivalent definitions for a mapping $f : X \to Y$ to be continuous at a point $a \in X$.

Let's look at an example. Let $X = Y = \mathbb{R}$ be the topological space the usual topology $\tau$ of open intervals. Let $f : \mathbb{R} \to \mathbb{R}$ be mapping defined for all $x \in X = \mathbb{R}$ by:

(2)

\begin{align} \quad f(x) = x + 1 \end{align}

We claim that $f$ is continuous at the point $0 \in X = \mathbb{R}$. To verify this, we must show that there exists local bases of $a = 0 \in X = \mathbb{R}$ and $f(a) = 1 \in Y = \mathbb{R}$, denoted then $\mathcal B_{a=0}$ and $\mathcal B_{f(a) = 1}$ such that for all $B \in \mathcal B_{f(a) = 1}$ there exists a $B' \in \mathcal B_{a=0}$ such that:

(3)

\begin{align} \quad f(B') \subseteq B \end{align}

Consider the following local bases of $a = 0$ and $f(a) = 1$:

(4)

\begin{align} \quad \mathcal B_{a=0} = \{ (a, b) : a, b \in \mathbb{R}, a < 0 < b \} \quad \mathcal B_{f(a) = 1} = \{ (c, d) : c, d \in \mathbb{R}, c < 1 < d \} \end{align}

Let $B \in \mathcal B_{f(a) = 1}$. Then $B = (c, d)$ where $c, d \in \mathbb{R}$ and $c < 1 < d$. Let $B' \in \mathcal B_{a=0}$ be defined by $B' = (a, b) = (c - 1, d - 1)$. Then we have that:

(5)

\begin{align} \quad f(B') = f((c-1, d-1)) = (c, d) \subseteq (c, d) = B \end{align}

Therefore $f$ is continuous at $0 \in \mathbb{R}$. In fact, it should not be too hard to see that $f$ is actually continuous for all $x \in X = \mathbb{R}$ using a similar argument for each $x \in X = \mathbb{R}$. Such mapping $f : X \to Y$ that are continuous for all $x \in X$ are said to be continuous (globally) which we define below.

Definition: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a mapping from $X$ to $Y$. Then $f$ is said to be Continuous or Continuous on All of $X$ if $f$ is continuous at every point $a \in X$.