GRE 9367 #62: Prove $X=[0,1]$ in lower limit topology ($[a,b)$) is not compact, is Hausdorff and is disconnected.

Question

GRE9367 #62

Ian Coley's solution:

Sean Sovine's solution:

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1. Prove $X$ is not compact.

My first proof was similar to Ian Coley's, but I came up with another proof:

If $X$ is compact, then because $X$ is Hausdorff, $X$ is compact Hausdorff in both standard and lower limit topologies of $\mathbb R$. This implies that the topologies are equal by (*), a contradiction.

Did I go wrong somewhere?

2. Prove $X$ is Hausdorff.

My proof is similar to Sean Sovine's. For Ian Coley's proof, is my understanding right?

If there exists the required open sets in standard topology, then we can choose the same sets as the required open sets in the lower limit topology.

3. Prove $X$ is disconnected.

My proof is the same as Ian Coley's. Is Ian Coley's proof right?

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(*) Munkres Exer26.1 (dbfin pf)

Answer 1

$[0,1]$ is not connected, e.g. because $[0,\frac12)$ is a non-trivial closed-and-open subset of it.

Not compact as $\{1-\frac1n: n=2,3,4,5,\ldots\}$ is an infinite subset without limit point in it.

Hausdorff because as you state its topology includes the usual, Hausdorff, one.

Answer 2

1. How did you get a contradiction? You can define two distinct Hausdorff compact topologies on the same set. But you can't if, furthermore, one of them is finer than the other one.
2. Yes, it is fine.
3. It is fine too.