This is an example from Munkres Topology book. In this example, it is shown that ℝ𝓁 has no countable basis. However it makes use of the axiom of choice. So I was wondering if it can be proved using a weaker version of choice. Here is such a proof:
Suppose for a contradiction that ℝ𝓁 is second countable.
Then since ℝ𝓁 is normal we can apply Urysohn's Lemma to conclude that ℝ𝓁 is metrizable.
Then ℝ𝓁×ℝ𝓁 is metrizable.
However ℝ𝓁×ℝ𝓁 is not normal which contradicts it being metrizable.
Therefore ℝ𝓁 is not second countable.
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In step 2 we used that ℝ𝓁 is normal. This fact can be proved without using any form of choice.
Similarly in step 4, it can be proved that ℝ𝓁×ℝ𝓁 is not normal using only the axiom of Dependant Choice[in The Baire category theorem form].
Urysohn's Lemma only requires the use of the axiom of Dependant Choice.
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Is there a more direct proof that doesn't use AoC?
