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I'd appreciate some feedback on whether or not I'm thinking about this problem correctly.

Show whether or not the subsets $Y \subset [0,1]\times[0,1] = X$ are open in $X$ when $X$ has the dictionary order topology. If $Y$ is open in $X$, express $Y$ as a union of basis elements. If $Y$ is not open in $X$, show that some point in $Y$ does not have a neighborhood in $Y$.

a) $(0,1)\times(0,1) \subset [0,1]\times[0,1]$
b) $(0,1)\times[0,1] \subset [0,1]\times[0,1]$
c) $[0,1]\times(0,1) \subset [0,1]\times[0,1]$

Since our set $X = [0,1]\times[0,1]$ has a least element $[0,0]$ and a greatest element $[1,1]$, then a basis $\mathcal B$ for the order topology on $X$ is $$\mathcal B = \{(x,y): x,y \in X, x < y\} \cup \{[0,y): y \in X\}\cup \{(x,1]: x \in X\}.$$

Thus,

a) $(0,1)\times(0,1)$ = $(0\times0, 1\times1)$

b) $(0,1)\times[0,1]$

not open since the smallest coordinate $0\times0$ does not have a neighborhood in $Y$, since the
x-coordinate of $0\times0$ is not in $Y$. Also, the largest coordinate $1\times1$ does not have a neighborhood in $Y$, since the $x$-coordinate of $1\times1$ is not in $Y$.

c) $[0,1]\times(0,1)$

not open since the smallest coordinate $0\times0$ does not have a neighborhood in $Y$, since the
$y$-coordinate of $0\times0$ is not in $Y$. Also, the largest coordinate $1\times1$ does not have a neighborhood in $Y$, since the $y$-coordinate of $1\times1$ is not in $Y$.

Teddy38
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user92638
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1 Answers1

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It looks to me as if you have a fundamental misunderstanding of some kind. In your answer to (b) you write:

Also, the largest coordinate 1 x 1 does not have a neighborhood in Y, since the x-coordinate of 1 x 1 is not in Y.

This doesn’t make sense: $Y$ is a set of ordered pairs of real numbers, so individual coordinates of those ordered pairs can’t possibly be elements of $Y$. In any case the point $1\times 1$, or $\langle 1,1\rangle$ as I prefer to write it, isn’t in $Y=(0,1)\times[0,1]$ anyway, since its first coordinate isn’t in $(0,1)$, so whether $\langle 1,1\rangle$ has an open nbhd in $Y$ is irrelevant.

As it happens, $(0,1)\times[0,1]$ is open. For each $x\in\left(0,\frac12\right)$ try to sketch the open interval

$$\left(\left\langle x,\frac12\right\rangle,\left\langle 1-x,\frac12\right\rangle\right)\;,$$

or in your notation

$$\left(x\times\frac12,(1-x)\times\frac12\right)\;;$$

can you show that every point of $Y$ is in one of these open intervals, and that each interval is a subset of $Y$?

In (a) the set $Y=(0,1)\times(0,1)$ is indeed open, but it’s not equal to $(0\times 0,1\times 1)$: $(0\times 0,1\times 1)$ is all of $X$ except the two endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$. HINT: Note that for each $x\in(0,1)$ the interval

$$\big(\langle x,0\rangle,\langle x,1\rangle\big)=(x\times 0,x\times 1)$$

is a subset of $Y$.

In (c) you’ve made the same error as in (b): it makes no sense to ask whether a coordinate of a point is in $Y$, and anyway the point $\langle 0,0\rangle$ isn’t in $Y$ in the first place. Do (a) first; once you’ve done it, this one should be pretty straightforward.

Brian M. Scott
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  • As it happens, (0,1)×[0,1] is open. For each $x \in (0,\frac12)$ try to sketch the open interval $(x$x$\frac12$, $(1-x)$x$\frac12)$; can you show that every point of Y is in one of these open intervals, and that each interval is a subset of Y?

    This is what my sketch of the open interval looks like. It doesn't seem to contain the points below $x$x$\frac12$ nor the points below $(1-x)$x$\frac12$. Isn't that problematic if I want to show that every point in Y is contained in this open interval? link

     – user92638 Oct 11 '14 at 20:09
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    @user92638: That’s not quite right: at the righthand end you should have points below $\langle 1-x,0.5\rangle$, not the points above it. – Brian M. Scott Oct 11 '14 at 20:11
  • This is the fixed version link. Why is it okay to ignore the points below $x$x$\frac12$ and the points above $(1-x)$x$\frac12$? Also, the open interval you gave for (0,1)x(0,1) is closed in [0,1]x[0,1], according to my lecture notes. Why does having the boundary points make your interval closed? Can't you choose not to include them, even though they're there? – user92638 Oct 11 '14 at 20:51
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    @user92638: The fixed version looks good. Consider a point like $x\times\frac14$ that’s below $x\times\frac12$: the first coordinates of $x\times\frac14$ and $x\times\frac12$ are equal, and $\frac14<\frac12$, so by the definition of dictionary order we have $x\times\frac14<x\times\frac12$, and therefore $x\times\frac14$ isn’t in the interval $(x\times\frac14,x\times\frac14)$. I don’t understand your last question: the interval $(x\times 0,x\times 1)$ is open in the dict. order on $[0,1]\times[0,1]$. – Brian M. Scott Oct 12 '14 at 03:02