A compactification of a topological space $X$ is a compact topological space $Y$ which contains a dense subspace homeomorphic to $X$. Let $X = (0,1]$, in the subspace topology of $\mathbb{R}$ and $f : X \longrightarrow \mathbb{R},\ x \mapsto \sin\frac{1}{x}$. Show the following: (A) $Y := [0,1]$ is a compactification of $X$, but $f$ does not extend to a continuous function $Y \longrightarrow \mathbb{R}$, i.e., there does not exist a continuous function $g : Y \longrightarrow \mathbb{R}$ such that $\left.g\right|_X = f$. (B) $X$ is homeomorphic to the set $X_1 := \left\{\left.\left(t, \sin\frac{1}{t}\right)\right\rvert\, t \in X\right\} \subseteq \mathbb{R}^2$. (C) The closure $Y_1$ of $X_1$ in $\mathbb{R}^2$ is a compactification of $X$. (D) $f$ extends to a continuous function $Y_1 \longrightarrow \mathbb{R}$.
A compactification of a topological space $X$ is a compact topological space $Y$ which contains a dense subspace homeomorphic to $X$. Let $X = (0,1]$, in the subspace topology of $\mathbb{R}$ and $f : X \longrightarrow \mathbb{R},\ x \mapsto \sin\frac{1}{x}$. Show the following:\\
(A) $Y := [0,1]$ is a compactification of $X$, but $f$ does not extend to a continuous function $Y \longrightarrow \mathbb{R}$, i.e., there does not exist a continuous function $g : Y \longrightarrow \mathbb{R}$ such that $\left.g\right|_X = f$.\\
(B) $X$ is homeomorphic to the set $X_1 := \left\{\left.\left(t, \sin\frac{1}{t}\right)\right\rvert\, t \in X\right\} \subseteq \mathbb{R}^2$.\\
(C) The closure $Y_1$ of $X_1$ in $\mathbb{R}^2$ is a compactification of $X$.\\
(D) $f$ extends to a continuous function $Y_1 \longrightarrow \mathbb{R}$.