Constant Mapping is Continuous

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Theorem

Let $T_A = \left({A, \tau_A}\right)$ and $T_B = \left({B, \tau_B}\right)$ be topological spaces.

Let $b \in B$ be any point in $B$.

Let $f_b: A \to B$ be the constant mapping defined by:

$\forall x \in A: f_b \left({x}\right) = b$


Then $f_b$ is continuous.


Proof

We have by definition that:

$\forall x \in A: f_b \left({x}\right) = b$

So: $f^{-1} \left({b}\right) = A$

For $c \in B: c \ne B$ we have that:

$f^{-1} \left({c}\right) = \varnothing$


Let $U \in \tau_B$ such that $b \in U$.

Then $f^{-1} \left({U}\right) = A$


Let $V \in \tau_B$ such that $b \notin V$.

Then $f^{-1} \left({V}\right) = \varnothing$


From the definition of topology, $A$ is open in $T$.

From Empty Set is Element of Topology, $\varnothing$ is also open in $T$.


So $f_b$ fulfils the conditions for it to be continuous.

$\blacksquare$


Sources