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From Wikipedia:

If the space $X$ is sequential, we may say that $x ∈ X$ is a limit point of a subset $S$ if and only if there is an $ω$-sequence of points in $S - \{x\}$ whose limit is $x$; hence, $x$ is called a limit point.

I was wondering how an $ω$-sequence of points is defined? Is $\omega$ a special ordinal?

How is the limit of an $ω$-sequence of points defined?

Thanks and regards!

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  • $\begingroup$ Yes, it is the first infinite ordinal. This means that the indices range over all ordinals $<\omega$, that is, the non-negative integers. $\endgroup$ – André Nicolas Jan 31 '12 at 5:30
  • $\begingroup$ Thanks! Then how is its limit defined? $\endgroup$ – Tim Jan 31 '12 at 5:31
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    $\begingroup$ $\omega$ represents the set of natural numbers $\mathbb N$. A $\omega$-sequence is just a fancy name for a usual sequence. $\endgroup$ – azarel Jan 31 '12 at 5:32
  • $\begingroup$ @azarel: Do you mean $\omega \equiv \mathbb{N}$? So a $\omega$-indexed sequence is a $\mathbb{N}$-indexed sequence, i.e. a sequence in the usual sense? $\endgroup$ – Tim Jan 31 '12 at 5:34
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    $\begingroup$ Tim, yes. $\omega$ sequence is a "regular" sequence. We make the distinction since often we want to allow longer sequences as well. $\endgroup$ – Asaf Karagila Jan 31 '12 at 5:39
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Yes, $\omega$ is the first infinite ordinal. This means that the indices range over all ordinals $<\omega$, that is, the non-negative integers, which are often defined as the finite ordinals.

A similar definition would work for any ordinal $\lambda$. The indices then range over all ordinals $<\lambda$.

For example, if $n$ is a non-negative integer, viewed as an ordinal, then an $n$-sequence has indices ranging over the $n$ (informal) ordinals $<n$, that is, over the ordinals $0,1,\dots,n-1$.

But we could let $\lambda$ be an ordinal $>\omega$. It seems to me you are unlikely to bump into them soon. You can view $\omega$-sequence as an excessively fancy name for an ordinary infinite sequence in the usual sense.

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