Path-connected and T1 with at least two points implies uncountable
(Redirected from Countable space with cofinite topology is not path-connected)
This article gives the statement, and possibly proof, of a topological space satisfying certain conditions (usually, a combination of separation and connectedness conditions) is uncountable.
Statement
The statement has the following equivalent forms:
- Any topological space that is both a path-connected space and a T1 space and has more than one point must be uncountable, i.e., its underlying set must have cardinality that is uncountably infinite.
- The closed unit interval cannot be expressed as a disjoint union of countably many non-empty closed subsets except in the trivial way (with only one piece.
- Any finite space or countable space with the cofinite topology is not path-connected. Note that the statement is obvious for finite spaces, so the chief content of the statement is that the countable space with cofinite topology is not path-connected.
![{\displaystyle [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
Related facts
- Connected and T1 with at least two points implies infinite
- Connected and regular with at least two points implies uncountable
- Connected and Urysohn with at least two points implies cardinality at least that of the continuum
- Connected and normal with at least two points implies cardinality at least that of the continuum
Proof
Proof of equivalence of the conditions
(1) and (3) are equivalent because the cofinite topology is the coarsest topology for a space to be T1, and path-connectedness is coarsening-preserved. The equivalence of (3) with (2) follows by noting that for a map from to be continuous to a space with cofinite topology is equivalent to the inverse images of all points in the space being closed subsets. Discarding the points not in the image of the map gives us our result.
Actual proof of (2)
The proof is a little lengthy, see the discussions here: