I How Does Closure in a Neighborhood Imply Membership in a Set?

[PsychonautQQ]

So I'm trying to understand a small part in the proof about how every 1-manifold is triangulable.

Let G be contained in K and let x be a limit point of G. Let U be a neighborhood of K that intersects G in finitely many closed neighborhoods, thus U intersect G is closed in G and thus x is in G.

Not undestanding:
U intersect G is closed in G implies x is in G

 

[WWGD]

So I'm trying to understand a small part in the proof about how every 1-manifold is triangulable.

Let G be contained in K and let x be a limit point of G. Let U be a neighborhood of K that intersects G in finitely many closed neighborhoods, thus U intersect G is closed in G and thus x is in G.

Not undestanding:
U intersect G is closed in G implies x is in G

It seems they are just using that a closed set contains all of its limit points. Assume otherwise. Then x is not in G, but it is a limit point. Notice that the complement of G is open ( in ambient space). Since x is a limit point of G , every 'hood $V_x$ of $x$ intersects the complement of G, which cannot happen, as the complement of G is open. Alternatively, if G is closed , so that its complement $G^c$ is open , and $x \in G^c$ then there is a hood $W_x$ of $x$ contained entirely in $W_x$. But this contradicts that $x$ is a limit point of $G$.

 

[PsychonautQQ]

does the intersection of the closure of U and G equal the intersection of the closure of G and the closure of U?

 

[WWGD]

does the intersection of the closure of U and G equal the intersection of the closure of G and the closure of U?

Not in every topological space. Let me see if it may apply on manifolds.

 

[PsychonautQQ]

It seems they are just using that a closed set contains all of its limit points.

So since x is a limit point of U and of G it must be a limit point of their intersection? Is this really obvious?

 

[WWGD]

So I'm trying to understand a small part in the proof about how every 1-manifold is triangulable.

Let G be contained in K and let x be a limit point of G. Let U be a neighborhood of K that intersects G in finitely many closed neighborhoods, thus U intersect G is closed in G and thus x is in G.

Not undestanding:
U intersect G is closed in G implies x is in G

Sorry for delay in replying. I think we are also using the fact that union of finitely many closed is closed: In subspace topology, each $K \cap G_i ; i=1,2,..,n$ (Each $G_i$ is one of the closed neighborhoods of intersection) is closed. Then we use that finite union of closed is closed.

 

[PsychonautQQ]

Sorry for delay in replying. I think we are also using the fact that union of finitely many closed is closed: In subspace topology, each $K \cap G_i ; i=1,2,..,n$ (Each $G_i$ is one of the closed neighborhoods of intersection) is closed. Then we use that finite union of closed is closed.

I must be really dense right now, but again, in the smallest words you can muster, why does this imply that x is in G?

 

[WWGD]

Don't worry, I am not seeing it that clearly myself. Is U supposed to contain G ? If so, G is the intersection of finitely-many closed sets and therefore closed, implying it contains all its limit points, in particular it contains x. Otherwise, not clear.

 

[PsychonautQQ]

Nope, U is not supposed to contain G. x a limit point of G, and U is a neighborhood of x, so by the definition of limit point we know that intersection(U,G) is not empty. We also know that this intersection is closed in U. Apparently this is supposed to imply that x is in G.

 

[WWGD]

Nope, U is not supposed to contain G. x a limit point of G, and U is a neighborhood of x, so by the definition of limit point we know that intersection(U,G) is not empty. We also know that this intersection is closed in U. Apparently this is supposed to imply that x is in G.

So U is ( U are? ;) ) a 'hood of both K and of x ? EDIT: If U is a 'hood of x , then the result is automatic.

 

[PsychonautQQ]

Well, U is a hood of x and this neighborhood is contained in K.

I think the main points are:
U is a neighborhood of x
Intersection(G,U) is closed in U
x is a limit point of G

How do these things imply that x is in G?

 

[WWGD]

Well, $U \cap G$ is closed and contained in G

Well, U is a hood of x and this neighborhood is contained in K.

I think the main points are:
U is a neighborhood of x
Intersection(G,U) is closed in U
x is a limit point of G

How do these things imply that x is in G?

Sorry, obviously I can't just grok it. Please give me some time, I will look at it more carefully.

 

[PsychonautQQ]

Lol no pressure dude you don't even have to help me if you don't want :P but thanks u r the best.