Talk:Topologist's sine curve

This page is useless without the figure. Could I scan it from a book? I guess a mathematical figure cannot be copyrighted. wshun 03:50, 9 Aug 2003 (UTC)

Image crashes browser when viewed full-size

Seems like this space is too weird for Iceweasel 2.0.0.3 (Debian version of Firefox) ;) Works fine in Konqueror. Functor salad 19:39, 19 July 2007 (UTC)

Something's very very wrong here

In the article, it says: You take the closure of the graph of sin(1/x) with x\in ]0,1]. The function is bounded, the domain is bounded, hence the graph is bounded. The closure of a bounded set w.r.t. the topology of a finite dimensional euclidean space is always compact. => Therefore, the topologist's sine curve is always compact, hence locally compact. But in the article, it says: "T is the continuous image of a locally compact space (namely, let V be the space {−1} ∪ (0, 1], and use the map f from V to T defined by f(−1) = (0,0) and f(x) = (x, sin(1/x)) for x > 0), but is not locally compact itself."

Later, it says in the article, that you may a variation, named "closed topologist's sine curve", which is now exactly the closure of the graph and therefore - by defintion - equal to the topologist's sine curve. So, the original topologist's sine curve is already the closed one...

I guess that some of the statements in this article refer to another sort of sine curve, where you just add (0,0) to the graph of sin(1/x). Then it would make sense to take the closure of it and then it would not be locally compact, but the image of a locally compact set (even a compact set)

The question is now: When topologists talk about "topologist's sine curve" do they mean the one with the interval or the one with just a point? --131.234.106.197 (talk) 16:42, 26 November 2008 (UTC)

Fixed. I don't know the answer to your last question. In Munkres, the closure is used. –Pomte 16:31, 11 December 2008 (UTC)