One-point compactification of $[1,2)\cup[3,4)\cup\{5\}$ I am asked to describe the one point compactification of X=$[1,2)\cup[3,4)\cup\{5\}$. I see that X is not a compact set, so X should be a  dense subset in $X_{+}$. It's seems that $[1,2]\cup[3,4)\cup\{5\}$ is homeomorphic to $X_{+}$. But $X_{+}$ is compact while $[1,2]\cup[3,4)\cup\{5\}$ isn't compact...
 A: Note that an isolated point remains such in the one-point compactification. Indeed, if $x$ is isolated in the locally compact space $X$ and $\infty$ is the added point: indeed $X\setminus\{x\}$ is a neighborhood of $\infty$.
Thus you only need to find the one-point compactification of $[1,2)\cup[3,4)$.
This is (homeomorphic to) any closed interval. Just reverse $[3,4)$ and identify the two terminal points.
A: A one point compactification does not require that the point be in $\mathbb R$ and behave like a point in $\mathbb R$.
It's not so much that you need to add a point-- you can add $2, 4, \infty$ or $-52$ or $\text{babar the elephant}$-- what matters is what the open sets you add.
If we add the point $\infty$ we must define an open set containing $\infty$ as $G\cup \{\infty\}$ there $G\subset [1,2)\cup [3,4)\cup \{5\}$, $G$ is open and $([1,2)\cup[3,4)\cup\{5\} )\setminus G$ is compact.
Those are just the sets $(2-\epsilon, 2)\cup (4-\delta,4)$.
So define sets of the form $(2-\epsilon,2)\cup (4-\delta,4)\cup \{\infty\}; 0< \epsilon < 1;0<\delta<1$ as open; and then $[1,3)\cup [3,4) \cup \{5\}\cup \infty$ is the one point campactification.
....
Alternatively.  You can define it as $[1,2]\cup [3,4)\cup\{5\}$ where $(2-\epsilon]\cup (4-\delta,4)$ is defined to be open.
Or as $[1,2]\cup [3,4]\cup\{5\}$ where $2$ is assumed to equal $4$ and the sets $(2-\epsilon, 2]\cup (4-\delta, 4]$ are assumedd to be open.
A: For $x\in [1,2)\cup \{5\}$ let $f(x)=x.$ For $x\in (3,4]$ let $f(x)=5-x.$ Then $f:X\to Y=[1,2)\cup (2,3]\cup \{5\}$ is a homeomorphism. Now $Y'=Y\cup \{2\}$ is a compactification of $Y.$ So let $X'=X\cup\{2\}$, and let $f(2)=2.$ And let  $U\subset X'$ be open in $X'$ iff $f[U]$ is open in $Y'.$
A: Let $Y$ = $[0,2] \cup \{3\}$ as a subspace of $\Bbb R$. This is compact Hausdorff.
Then note that $$[0,2] \setminus \{1\} = [0,1) \cup (1,2] \cup \{3\} \simeq [0,1) \cup [1,2) \cup \{3\} \simeq X$$ So $Y$ is homeomorphic to the one-point compactification of $X$, by the essential unicity of this compactification.
