Timeline for Is the empty graph a tree?
Current License: CC BY-SA 3.0
10 events
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| Feb 28, 2013 at 11:46 | comment | added | rgrig | In computer science, (ordered) binary trees are usually defined as a (least) fixed point of $1+T\times T=T$. The empty tree is a binary tree by definition. Binary trees seem to be rather different from connected acyclic graphs. | |
| Feb 26, 2013 at 22:49 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Feb 4, 2013 at 22:14 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Feb 3, 2013 at 16:23 | comment | added | Samuel Vidal | May be a way to clarify the question is to consider not the graph in itself but within its family. Consider The family (species) of connected graphs $G^c$ and the family of not necessarily connected graphs $G$. The relation between them is: $$G\simeq E(G^c)$$ where E stands for the species of sets. This natural isomorphism comes from existence and unicity of a decomposition of a graph (in $G$) in its connected components (in $G^c$). If you work out the details, $G^c$ can't have any graph of size zero. This is why you better not see the empty graph as connected. | |
| Feb 3, 2013 at 5:52 | comment | added | Aaron Meyerowitz | Of course. I'm just saying that one could decide to do so. As I said, it does seem a stretch and not of much obvious use. | |
| Feb 3, 2013 at 5:17 | comment | added | Todd Trimble | All the definitions of rooted tree I've seen (except in your answer) stipulate a distinguished node called the root. | |
| Feb 3, 2013 at 4:26 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Feb 3, 2013 at 3:43 | comment | added | Aaron Meyerowitz | Todd, that only holds if it has vertices! | |
| Feb 2, 2013 at 13:57 | comment | added | Todd Trimble | Well of course a rooted tree can't be empty, because it has a root. | |
| Feb 1, 2013 at 20:26 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |