Prove that the set is close

Question

Given metric spaces $(X,d)$ and $(Y,d')$ and continuous mapping $S$ and $T$ from $X$ into $Y$, prove that the set $\{x \in X: Sx = Tx\}$ is closed in $(X,d)$.

I've run out of any ideas where I should start

Answer 1

Let $E=\{x∈X:Sx=Tx\}$. Suppose $x_n \in E$ and $x_n \to x$.

By continuity, $S x_n \to Sx$ and $T x_n \to Tx$. But $Sx_n=Tx_n$ for all $n\ge1$, hence, $Sx=Tx$ and this means $x\in E$. This implies $E=\{x∈X:Sx=Tx\}$ is closed.

Answer 2

You can give a topological proof as follows:

Consider the function $D:X\rightarrow \mathbb{R}$ defined by $D(x)=d(S(x),T(x))$. Since $S$ and $T$ are continuous and the distance function $d$ is continuous, then $D$ is continuous, so given $F$ a closed subset in $\mathbb{R}$ we have that: $$D^{-1}(F)=\{x\in X:D(x)\in F\}$$ is a closed subset in $X$.

We also know that in a metric space, the points are closed. In particular, $\{0\}$ is a closed set in $\mathbb{R}$.

And we have finished, because: $$\{x\in X: S(x)=T(x)\}=D^{-1}(\{0\})$$ which is closed.