$X$ is said to have the universal extension property if for every normal space $Y$ and every closed subset $A \subset Y$ and every continuous function $f : A \longrightarrow X$, $f$ extends to a continuous function from $Y$ to $X$. You may assume, without proof, that $\mathbb{R}^{2}$ has the universal extension property. (A) Prove or find a counterexample: If $X$ has the universal extension property, then $X$ is connected. (B) Give an example (with justification) of a compact subset $X$ of $\mathbb{R}^{2}$ that does not have the universal extension property. (C) Let $X = \{(x, \sin x) \mid x \in \mathbb{R}\}$. Then show that $X$ has the universal extension property.
$X$ is said to have the universal extension property if for every normal space $Y$ and every closed subset $A \subset Y$ and every continuous function $f : A \longrightarrow X$, $f$ extends to a continuous function from $Y$ to $X$. You may assume, without proof, that $\mathbb{R}^{2}$ has the universal extension property.\\
(A) Prove or find a counterexample: If $X$ has the universal extension property, then $X$ is connected.\\
(B) Give an example (with justification) of a compact subset $X$ of $\mathbb{R}^{2}$ that does not have the universal extension property.\\
(C) Let $X = \{(x, \sin x) \mid x \in \mathbb{R}\}$. Then show that $X$ has the universal extension property.