Is the axiom of countable choice needed for the proof of this special case of equivalence of limit definitions?

Asked Feb 21 '16 at 19:24

Active Feb 21 '16 at 19:24

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In this link, a proof of the equivalence between the $\delta-\epsilon$ definition and the one with the convergent sequences is given.

I'm interested in the proof of this theorem simply for the metric space $(\Bbb R, |x-y|)$, and I was wondering if the axiom of countable choice was really needed in the sufficient condition part.

If there's a different proof of the general theorem which avoids this axiom, it would be very nice.

asked Feb 21 '16 at 19:24

YoTengoUnLCD

13,384

To sum up the two answers in the duplicate: If you are talking about continuity at a single point, yes; if you're talking about continuity everywhere, no. – Asaf Karagila Feb 21 '16 at 19:31
@AsafKaragila : Could you link the dupe? – Eric Towers Feb 21 '16 at 19:35
@Eric: It's on the banner on the top of the question... – Asaf Karagila Feb 21 '16 at 19:41
@AsafKaragila : I see no banner. (Also, SE seems very laggy today (here).) – Eric Towers Feb 21 '16 at 19:46
@EricTowers Here it is: http://math.stackexchange.com/questions/126010/continuity-and-the-axiom-of-choice .
Btw, the banner looks like this: http://i.imgur.com/i6p2kaf.png
– YoTengoUnLCD Feb 21 '16 at 19:48

Is the axiom of countable choice needed for the proof of this special case of equivalence of limit definitions?

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