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Noah Schweber
Noah Schweber
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If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable.

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just choose the lengths successively.

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just choose the lengths successively.

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable.

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just choose the lengths successively.

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Fedor Petrov
Fedor Petrov
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If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just wchoosechoose the lengths successively.

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just wchoose the lengths successively.

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just choose the lengths successively.

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Fedor Petrov
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

If the nodes add a fixed amount of new nodes at each level, then the number of infinite paths seems to be countable. -

It does not seem so for me. Even if the number of nodes $k_n$ on level $n$ satisfies $k_n\leqslant k_{n+1}\leqslant k_n+1$, the number of infinite paths may have cardinality continuum. For constructing an example take an infinite binary tree and replace each edge to a path of certain length, choosing lengths so that all branchings occur at different levels. This is possible, just wchoose the lengths successively.