Is every infinite compact space with no isolated points uncountable?
I know that every nonempty Hausdorff compact space with no isolated points is uncountable, so I was wondering if we could substitute the nonempty Hausdorff part with it being infinite.
general-topology
asked yesterday
Ryunaq
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I know that every nonempty Hausdorff compact space with no isolated points is uncountable, so I was wondering if we could substitute the nonempty Hausdorff part with it being infinite.
general-topology
asked yesterday
Ryunaq
1356
add a comment |Â
I know that every nonempty Hausdorff compact space with no isolated points is uncountable, so I was wondering if we could substitute the nonempty Hausdorff part with it being infinite.
general-topology
asked yesterday
Ryunaq
1356
I know that every nonempty Hausdorff compact space with no isolated points is uncountable, so I was wondering if we could substitute the nonempty Hausdorff part with it being infinite.
general-topology
general-topology
asked yesterday
Ryunaq
1356
asked yesterday
Ryunaq
1356
asked yesterday
Ryunaq
1356
asked yesterday
Ryunaq
1356
asked yesterday
Ryunaq
1356
1356
add a comment |Â
add a comment |Â
1 Answer
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No. For instance, you could take a countably infinite set with the indiscrete topology.
You could consider that example to be cheating, as its $T_0$ quotient does have isolated points (so the space has points which are "isolated" from all points that they are topologically distinguishable from at all). For an example which is additionally $T_0$ (even $T_1$), you could take a countably infinite set with the cofinite topology.
answered yesterday
Eric Wofsey
179k12204331
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No. For instance, you could take a countably infinite set with the indiscrete topology.
You could consider that example to be cheating, as its $T_0$ quotient does have isolated points (so the space has points which are "isolated" from all points that they are topologically distinguishable from at all). For an example which is additionally $T_0$ (even $T_1$), you could take a countably infinite set with the cofinite topology.
answered yesterday
Eric Wofsey
179k12204331
add a comment |Â
No. For instance, you could take a countably infinite set with the indiscrete topology.
You could consider that example to be cheating, as its $T_0$ quotient does have isolated points (so the space has points which are "isolated" from all points that they are topologically distinguishable from at all). For an example which is additionally $T_0$ (even $T_1$), you could take a countably infinite set with the cofinite topology.
answered yesterday
Eric Wofsey
179k12204331
add a comment |Â
No. For instance, you could take a countably infinite set with the indiscrete topology.
You could consider that example to be cheating, as its $T_0$ quotient does have isolated points (so the space has points which are "isolated" from all points that they are topologically distinguishable from at all). For an example which is additionally $T_0$ (even $T_1$), you could take a countably infinite set with the cofinite topology.
answered yesterday
Eric Wofsey
179k12204331
No. For instance, you could take a countably infinite set with the indiscrete topology.
You could consider that example to be cheating, as its $T_0$ quotient does have isolated points (so the space has points which are "isolated" from all points that they are topologically distinguishable from at all). For an example which is additionally $T_0$ (even $T_1$), you could take a countably infinite set with the cofinite topology.
answered yesterday
Eric Wofsey
179k12204331
answered yesterday
Eric Wofsey
179k12204331
answered yesterday
Eric Wofsey
179k12204331
answered yesterday
Eric Wofsey
179k12204331
179k12204331
add a comment |Â
add a comment |Â
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