# Open sets in $\mathbb R\times \mathbb R$ with the dictionary order

In the example below, I wonder why Munkres depicted the first type of intervals in this way?

More specifically, why doesn't the picture look like this:

where all points below, above, and on the red segment are removed?

This is my justification why I think this should be the right picture:

$$(a\times b,c\times d)=\{x|a \times b <_{dict} x <_{dict} c\times d\}\\=\{(p,q)|a <_{R} p <_{R} c\}\cup \{(r,s)| a=r=c \text{ and } c <_{R} s <_{R} d \}$$ (where $<_{dict}$ is the dictionary order and $<_R$ is the standard order on $\mathbb R$). The first set in the union is the set of the first type, the second set in the union is the set of the second type.

Now, any point above, below, or on the red line has the first coordinate $a$, which cannot lie in a set of the first type.

Update: I defined the sets of the first type in a way that differs from Munkres, but I don't see what's wrong with my description of open sets (in the display), and no open set in my description can be a set of the first type in Munkres.

• Um, removing the red segments on your second image produces exactly the first one, doesn't it? – Henning Makholm Sep 1 '18 at 19:12
• @HenningMakholm I'm not removing just the red segment, I'm removing everything above and below it as well. – user531587 Sep 1 '18 at 19:21
• What then happen to points such as $a\times(b+1)$, which is definitely both $>_{\rm dict} a\times b$ and $<_{\rm dict} c\times d$? – Henning Makholm Sep 1 '18 at 19:29
• @HenningMakholm okay, this makes sense... I was confused (and still confused) by the description of the basis in the text. – user531587 Sep 1 '18 at 19:51
• Actually this answer made things more clear: math.stackexchange.com/a/1129061/531587 – user531587 Sep 1 '18 at 20:02

Suppose a point $p = (x,y) \in (a \times b , c \times d)$ has $a$ as the first coordinate. Since $(a,b) < (x,y) = (a,y)$, this necessarily implies that $b < y$. In the same vein, if $x = c$ and since $(x,y) = (c,y) < (c,d)$, necessarily we must have that $y < d$. This justifies why the interval only contains an open ray of $\{x = a\}$ and $\{x = c\}$ instead of the whole line.
Your characterization of the interval is not quite right. Note that $r$ cannot be $a$ and $c$ at the same time, so you should separate in two cases (which is basically what I wrote) depending on the value that the first coordinate takes. This should clarify the picture.