Show that $\psi_v:\mathbb{K}\rightarrow\mathbb{K}\times X $ given by $\psi_v(\xi)=(\xi,v)$ is continuous

Lemma. Let $X$ be a topological vector space over $\mathbb{K}$ and $v\in X$. Then the following mapping is continuous $$\begin{align*} \varphi _v:\mathbb{K}&\rightarrow X \\ \xi &\mapsto \xi v. \end{align*}$$

Proof. For any $\xi\in\mathbb{K}$, we have $\varphi_v(\xi)=M(\psi _v\left ( \xi \right ))$, where $\psi_v:\mathbb{K}\rightarrow\mathbb{K}\times X $ given by $\psi_v(\xi)=(\xi,v)$ is clearly continuous by definition of product topology and $M:\mathbb{K}\times X$ is the scalar multiplication in the topological vector space. $X$ which is continuous by definition of topological vector space. Hence, $\varphi_v$ is continuous as composition of continuous mappings.

My question is:

• Why $\psi_v$ is continuous by definition of product topology? I have looked for the definition here Product topology but there is no information that I need.

Can someone help me? Thanks.

Solution 1:

Let $A,B, C$ be topological spaces. A function $\phi : A \to B \times C$ is continuous iff the functions $p_B \circ \phi : A \to B$ and $p_B \circ \phi : A \to C$ are continuous, where $p_B : B \times C \to B$ and $p_C : B \times C \to C$ are the projection maps. This is known as the universal property of the product topology on $B \times C$.

We have $p_{\mathbb K} \circ \psi_v = id_{\mathbb K}$ which is continuous. Moreover $p_X \circ \psi_v$ is the constant map with value $v$ which is also continuous.