I am confused on the following theorem. Let X be a space. Then X is locally compact Hausdorff if and only if there exists a space Y satisfying the following conditions:
(1) X is a subspace of Y
(2) The set $Y-X$ consists of a single point.
(3) Y is a compact Hausdorff space.
Consider $\mathbb{R}$ under the usual topology and the subspace topology on $(0,1)$. The interval $(0,1)$ is locally compact, however the compactification requires two points, namely $0$ and $1$. Therefore $Y – (0,1)$ will consist of 2 points not one. I am very confused, where did I go wrong?
Also, I am having trouble showing that the compactification of $\mathbb{R}$ is homomorphic to $S_{1}$. I can show that the circle is compact and Hausdorff under the supspace topology of $\mathbb{R}^2$ where $\mathbb{R}^2$ is equipped with the topology that is induced by the Euclidean metric. But I need a continuous function from $\mathbb{R}$ into the circle.
Best Answer
Topologically, the standard open interval is the same as the real line. To see this, use the maps $$(-1..1) → ℝ,~x ↦ \frac x {1 - |x|},\quad\text{and}\quad ℝ → (-1..1),~x ↦ \frac x {1 + |x|},$$ and note that $(0..1) \cong (-1..1)$ by some affine transformation.
Of course $[0..1]$ is also a compactification of $(0..1)$ – if you understand a “compactification” of a space to merely be a compact space in which the original space can be embedded. But then it’s not the only one and it’s not the one-point compactification. In general, there are many ways to compactify a space in this sense.
In fact, $(0..1)$ can be compactified to a circle. To see this, just think about how you would create a loop out of a loose string and some glue. If you know about trigonometric functions, or better yet, the complex exponential, you’ll see how to do this mathematically.
If you want to compactify $ℝ$ directly, you can go for the stereographic projection.