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What does $0\times 1$ mean in the order topology $?$ How does ${{1}\over{2}} \times 0$ look like? Are they just a point or a line$?$ How do i visualize them$?$ I understand that $[0,1]\times[0,1]$ is a square created by the interval $[0,1]$ on the $x$-axis (that is on $y=0$) that is being extended vertically up to $y=1$. Is my understanding right$?$

user118494
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desperatemuch
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1 Answers1

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I’m going to guess that you’re reading Munkres. Munkres, unfortunately, uses a non-standard notation for ordered pairs: his $0\times 1$ is the point whose $x$-coordinate is $0$ and whose $y$-coordinate is $1$. I would write it $\langle 0,1\rangle$, though you’re probably more familiar with the notation $(0,1)$. Thus, it’s the upper left corner of the square $[0,1]\times[0,1]$, which you do understand correctly.

Similarly his $\frac12\times 0$ is the point that I would write $\left\langle\frac12,0\right\rangle$ and that most calculus texts, for instance, would write $\left(\frac12,0\right)$: it’s the point in the middle of the bottom edge of the unit square $[0,1]\times[0,1]$.

Brian M. Scott
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  • Thank yo Brian M. Scott. It is clear to me now. I just have a follow up question regarding a set A={1/n \times 0: n is a natural number}. I understand that 1/n goes to zero but what confuses me is how this set is being represented geometrically. Are these the set of all points (1/n, 0)? If so, then I do not quite understand why according to my teacher the closure of A is the union of A and {0 \times 1} – desperatemuch Aug 31 '15 at 07:45
  • @desperatemuch: You’re welcome. – Brian M. Scott Aug 31 '15 at 07:45
  • I would like to explain my confusion, why wouldn't it be the union of A and {0 x 0}? – desperatemuch Aug 31 '15 at 07:52
  • @desperatemuch: Yes, $A$ is the set of points $\left\langle\frac1n,0\right\rangle$ for positive integers $n$. The closure of $A$ would be $A\cup{\langle 0,0\rangle}$ if you were working in the Euclidean topology on $[0,1]\times[0,1]$; I’m guessing from your teacher’s statement that you’re working in $[0,1]\times[0,1]$ with the lexicographic order topology. In that case the set ${0}\times[0,1)$ is an open nbhd of $\langle 0,0\rangle$ that is disjoint from $A$. However, every open nbhd of $\langle 0,1\rangle$ in the lex. order top. contains as a subset a set of the form ... – Brian M. Scott Aug 31 '15 at 07:55
  • ... $(0,\epsilon)\times[0,1]$ for some $\epsilon>0$ and therefore contains infinitely many points of $A$. – Brian M. Scott Aug 31 '15 at 07:56
  • oh my. so how does {0}x[0,1] look like? Am I doing this right? I am desperately trying to visualize these sets? or am I totally lost? Would {0}x[0,1) be intervals (0,t) where t is in [0,1)? – desperatemuch Aug 31 '15 at 08:01
  • @desperatemuch: ${0}\times[0,1)$ is all of the lefthand edge of the square except the point at the very top. – Brian M. Scott Aug 31 '15 at 08:02
  • or {0} x [0,1] just like a vertical line on the y axis from y=0 to y=1? And that (0,\epsilon) \times [0,1] is a neighborhood/square ? – desperatemuch Aug 31 '15 at 08:05
  • @desperatemuch: Yes to both questions (except that $(0,\epsilon)\times[0,1]$ is a rectangle, not a square). You might find my answers to this question, this question, and this question helpful in seeing more of what’s going on. – Brian M. Scott Aug 31 '15 at 08:08
  • Another question. I hope This will be the final one, @Brian M. Scott.. I understand that a closure of a set is made from the union of the set itself and its limit points. Hence, in looking for the limit point(s) of sets with the lexicographic order, should i try to identify first what its limit point in the the Euclidean topology and try to get its corresponding open neighborhoods (with respect to the lexicographic order topology)? – desperatemuch Aug 31 '15 at 08:11
  • @desperatemuch: The Euclidean topology isn’t really of much use except along the vertical lines through the square: the topologies are simply too different. I would start by making sketches, similar to the ones in the second link that I posted, of typical open nbhds of points of the lex. ord. square. You three basic types: points that are not on the top or bottom edge, points that are on the top edge, and points that are on the bottom edge. (The lower left and upper right corner points are slightly special cases.) Once you know what the basic open nbhds of points look like, it’s much ... – Brian M. Scott Aug 31 '15 at 08:15
  • ... easier to see which points are good candidates to be limit points of a given set. (You’re welcome.) – Brian M. Scott Aug 31 '15 at 08:15
  • Hi, I am not allowed to post a a follow up question on the second link that you gave. My mind is still cloudy with the representation of the open interval (a_n, b_n). Why is the sketch colored orange all the way up? why didn't it end with y=b_n? – desperatemuch Aug 31 '15 at 08:43
  • @desperatemuch: Because every one of those orange points lies between the two blue points in the lexicographic order. Remember, $\langle a,b\rangle$ precedes $\langle c,d\rangle$ in the lex. order whenever $a<c$, no matter what $c$ and $d$ are; you only have to worry about $c$ and $d$ when $a=b$, in which case $\langle a,b\rangle$ precedes $\langle c,d\rangle$ if $c<d$. (It’s almost 5 a.m. here, and I’m about to go get some sleep, but I’ll check back in a few hours.) – Brian M. Scott Aug 31 '15 at 08:47
  • This question and answer has sparked a little self-awareness: since learning topology from Munrkes almost 40 years ago, I have also continued to occasionally use the notation $x \times y \in X \times Y$ as an alternate for the ordered pair notation $(x,y) \in X \times Y$. – Lee Mosher Aug 31 '15 at 12:55