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Piotr Hajlasz
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I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin, Combinatorial Optimization in Geometry Adv. in Appl. Math. 31 (there2003), no. 1, 242–271. (There is a free arxiv.org versionfree arxiv.org version, in case that matters).)

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of I. Rivin, Combinatorial Optimization in Geometry Adv. in Appl. Math. 31 (2003), no. 1, 242–271. (There is a free arxiv.org version, in case that matters.)

replaced http://mathoverflow.net/ with https://mathoverflow.net/
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I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

added argument outline
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Igor Rivin
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I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

I am not sure which direction gives you trouble, but the second condition is necessary, by the Moore-Young theorem, or a somewhat weaker version thereof, as discussed by Greg Kuperberg here.

For the right direction, condition 2 says that there are countably many vertices, since you can just erase those of degree 2, and vertices of degree 1 clearly don'make any difference, planarity wise. So, now, we have a graph with countably many vertices, and thus countably many edges, such that any finite subgraph is planar. You want to show that the graph is planar. This is a simple compactness/Arzela-Ascoli type argument - something very similar (but stronger) is showed in the last section of "Combinatorial Optimization in Geometry" by one I. Rivin (there is a free arxiv.org version, in case that matters).

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Igor Rivin
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