I need to prove that every locally Euclidean space is actually a $T_1$ space but I do even know how to start. Any help that you bring to me will be appreciated...
Asked
Active
Viewed 301 times
1
-
What would happen if there were two distinct points that could not be separated by any open sets? Remember, we can always find a neighborhood around any point that is homeomorphic to an open set in $\mathbb R^n$. – John Douma Mar 10 '19 at 22:19
-
@JohnDouma a locally Euclidean space need not be Hausdorff. – Henno Brandsma Mar 10 '19 at 22:23
-
@HennoBrandsma If we took an open neighborhood $U$ around $x$ that was homeomorphic to $\mathbb R^n$ and also contained $y$, couldn't we then get two open subsets of $U$ that separated $x$ and $y$, yielding a contradiction? – John Douma Mar 10 '19 at 22:27
-
@JohnDouma no, look up the line with two origins, e.g. – Henno Brandsma Mar 10 '19 at 22:28
-
@HennoBrandsma Yes, thank you. That's because the basis elements that contain each origin also contain intervals which will always intersect. Thus, even though $O_1\not\in (-a,0)\cup {O_2}\cup (0,a)$ the two basis elements always intersect. – John Douma Mar 10 '19 at 22:46
-
@JohnDouma Hausdorff is not a "local enough" property, and $T_1$ is. It's a nice example (a standard one in topological manifold theory). – Henno Brandsma Mar 10 '19 at 22:49
1 Answers
4
Let $x \neq y$ in $X$. Let $U$ be an open Euclidean neighbourhood of $x$. If $y \notin U$ we are done. If not, in the Euclidean (hence $T$-whatever) subspace $U$ we can find an open neighbourhood $V$ of $x$ that misses $y$. As $V$ is still open in $X$ ("open in open is open"), the same $V$ witnesses $T_1$-ness for $x,y$.
In short, a locally $T_1$ (or $T_0$) space is globally $T_1$ (resp. $T_0$).
The same argument shows that a locally $R_0$ space is globally $R_0$, where an $R_0$ space is one where, for any two topologically distinguishable points, each has a neighbourhood not containing the other point.
PatrickR
- 7,165
Henno Brandsma
- 250,824