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For every $(a,b)\in \mathbb{R}^2$ and $\alpha,\beta$ positive reals, define $$J_{a,b,\alpha,\beta}=\{(x,y)\in \mathbb{R}^2 \mid a<x<a+\alpha \; \mathrm{and} \; b<y<b+\beta \}\cup \{ (a,b)\}$$

$(1)$ Show that $\mathfrak{B}:= \{J_{a,b,\alpha,\beta} :(a,b)\in \mathbb{R}^2 \; \mathrm{and}\; \alpha,\beta \in \mathrm{R}^+ \}$ $(2)$ Is $\tau$ comparable with the euclidean topology in $\mathbb{R}^2$? Is $(\mathbb{R}^2,\tau)$ Hausdorff?

$(3)$ Find the closure of $J_{a,b,\alpha,\beta}$. Show that it is not compact. Is $(\mathbb{R}^2,\tau)$ locally compact?

$(4)$ Find the induced topology on the diagonal $\Delta :=\{ (x,x)\mid x\in \mathbb{R}\}$ and in $\Omega:=\{(x,-x)\mid x\in \mathbb{R}\}$

$(5)$ Find an open and closed subset, distinct to the whole space and the empty space. Show that the connected components of $(\mathbb{R}^2,\tau)$ are unipuntual.

My try:

For $(1)$, for every $(x,y)\in \mathbb{R}^2$, $(x,y)\in J_{x,y,\alpha,\beta}$ and if $(x,y)\in J_{a_1,b_1,\alpha_1,\beta_1}\cap J_{a_2,b_2,\alpha_2,\beta_2}$, then let $$a=\min \{a_1,a_2\},\; b=\min \{b_1,b_2\},\; \alpha=\min\{\alpha_1,\alpha_2\},\;\beta= \min\{\beta_1,\beta_2\}$$ Then $(x,y)\in J_{a,b,\alpha,\beta} \subset J_{a_1,b_1,\alpha_1,\beta_1}\cap J_{a_2,b_2,\alpha_2,\beta_2}$, so $\mathfrak{B}$ is a base of a topology $\tau$

For $(2)$, to compare $\tau$ with $\tau_u$, let $B_r^a$ be any open ball with center $a$ and radius $r$, then for every $(x,y)\in B_r^a$ we can find $\alpha$ and $\beta$ small enough such that$J_{x,y,\alpha,\beta}\subset B_r^a$.

On the other hand, let $J:=J_{a,b,\alpha,\beta}$ and take the point $(a,b)$. Then, for every $r>0$, $B((a,b),r)$ is not contained in $J$. So $$\tau_u \subset \tau$$ But $$\tau \not \subset \tau_u$$ Now, to prove that the space is Hausdorff, take two distinct points $(x_1,y_1),(x_2,y_2)$ and assume $x_1<x_2$ and $y_1<y_2$. Fix $a_1$ and $\alpha_1$ such that $$a_1<x_1<a_1+\alpha_1 <x_2$$ and find $a_2$ and $\alpha_2$ such that $a_1+\alpha_1 <a_2<x_2<a_2+\alpha_2$. Now repite the same argument for $y_1$ and $y_2$ and we'd find $J_1=J_{a_1,b_1,\alpha_1,\beta_1}$ and $J_2=J_{a_2,b_2,\alpha_2,\beta_2}$ such that $(x_1,y_1)\in J_1$, $(x_2,y_2)\in J_2$ and $J_1\cap J_2= \emptyset$, so $(\mathbb{R}^2,\tau)$ is Hausdorff.

For $(3)$, I have a vague idea of what to prove but I'm stuck in proving some of the parts. I strongly believe that the closure of $J_{a,b,\alpha,\beta}$ is $$\overline{J}_{ a,b,\alpha,\beta}=\{(x,y)\in \mathbb{R}^2 \mid a\leq a+\alpha,\; b\leq y \leq b+\beta \}$$ but I'm not sure how to prove it or if it is even correct. To prove that the closure is not compact I have no idea. I know that a compact subset of a Hausdorff space is closed, but that doesn't help since the closure is always a closed subset.

For $(4)$ I'm not sure how to find the induced topology. I think it depends if $(x,y)\in \Delta$ or not. If $(x,y)\in \Delta$, then the subspace topology would be the lower limit topology in $\mathbb{R}$, and if $(x,y)$ is not in $\Delta$, then it would be the usual topology $\tau_u$ of $\mathbb{R}$, and the same would apply to $\Omega$, changing the lower limit topology for the upper limit topology instead.

For the last part, $(5)$ I really do't have a clue.

I'd be grateful if somebody could help me out with this giving me a hint for the parts I haven't answered yet or correcting me in the parts I have answered. Thanks in advance!

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  • $\begingroup$ @Aweygan It was a typo, the singleton should be $(a,b)$ instead. $\endgroup$ – user313212 Aug 20 '16 at 14:07
  • $\begingroup$ Woah hoho, a downvote already? $\endgroup$ – IAmNoOne Aug 20 '16 at 14:08
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HINT: I assume that $(1)$ was supposed to be to prove that $\mathfrak{B}$ is a base for a topology on $\Bbb R^2$. Your argument doesn’t quite work, because your $J_{a,b,\alpha,\beta}$ isn’t necessarily contained in the intersection of $J_{a_1,b_1,\alpha_1,\beta_1}$ and $J_{a_2,b_2,\alpha_2,\beta_2}$. For example, $J_{0,0,2,2}\nsubseteq J_{0,0,2,2}\cap J_{1,1,2,2}$.

Suppose that $\langle x,y\rangle\in J_{a_1,b_1,\alpha_1,\alpha_2}\cap J_{a_2,b_2,\alpha_2,\beta_2}$. Then $x<\min\{a_1+\alpha_1,a_2+\alpha_2\}$, so let

$$\alpha=\min\{a_1+\alpha_1,a_2+\alpha_2\}-x\;.$$

Similarly, $y<\min\{b_1+\beta_1,b_2+\beta_2\}$, so let

$$\beta=\min\{b_1+\beta_1,b_2+\beta_2\}-y\;.$$

Now check that $\langle x,y\rangle\in J_{x,y,\alpha,\beta}\subseteq J_{a_1,b_1,\alpha_1,\alpha_2}\cap J_{a_2,b_2,\alpha_2,\beta_2}$.

Your proof that $\tau_u\subsetneqq\tau$ is correct. Your proof that $\tau$ is Hausdorff, however, is incomplete, because you considered only the case $x_1<x_2$ and $y_1<y_2$. You need to cover all cases. The easiest way is to note that if $x_1<x_2$, then $J_{x_1,y_1,x_2-x_1,1}$ and $J_{x_2,y_2,1,1}$ are disjoint open nbhds of $\langle x_1,y_1\rangle$ and $\langle x_2,y_2\rangle$, while if $y_1<y_2$, we can use $J_{x_1,y_1,1,y_2-y_1}$ and $J_{x_2,y_2,1,1}$. Without loss of generality one of these must be the case.

For $(3)$, your guess at the closure of $J_{a,b,\alpha,\beta}$ isn’t quite right. Notice that if we set $c=a+\alpha$ and $d=b+\beta$, then

$$J_{a,b,\alpha,\beta}=\{\langle a,b\rangle\}\cup\big((a,c)\times(b,d)\big)\;.$$

This is an open box in $\Bbb R^2$ together with its lower left corner. It’s not hard to see that a point on the left edge of the box (other than the top point on that edge) is in the closure, as a point on the bottom edge (other than the righthand endpoint of that edge). Points along the top and righthand edges, however, all have open nbhds disjoint from the box. Thus, you should try to prove that

$$\begin{align*} \operatorname{cl}J_{a,b,\alpha,\beta}&=[a,c)\times[b,d)\\ &=\{\langle x,y\rangle\in\Bbb R^2:a\le x<a+\alpha\text{ and }b\le y<b+\beta\}\;. \end{align*}$$

Do this by showing that if $b<y<b+\beta$, then every open nbhd of $\langle a,y\rangle$ intersects $J_{a,b,\alpha,\beta}$, and if $a<x<a+\alpha$, then every open nbhd of $\langle x,b\rangle$ intersects $J_{a,b,\alpha,\beta}$, while if $x\ge a+\alpha$ or $y\ge b+\beta$, then $\langle x,y\rangle$ has an open nbhd disjoint from $J_{a,b,\alpha,\beta}$.

To deal with the compactness parts of $(3)$, show that for any real number $\gamma$, every subset of the line $x+y=\gamma$ is a closed, discrete set in this topology. Since the closure of every member of $\mathfrak{B}$ contains an infinite subset of such a line, none of these closures can be compact: a compact set cannot contain an infinite closed discrete subset. (Why?)

For $(4)$, use the previous remark to show that $\Omega$ inherits the discrete topology from $\langle\Bbb R^2,\tau\rangle$: for each $p\in\Omega$ there is a $J\in\mathfrak{B}$ such that $J\cap\Omega=\{p\}$. Then show that $\Delta$ with its relative topology is homeomorphic to the Sorgenfrey line (which you may know as the real line with the lower limit topology). Think about what $J_{x,x,\alpha,\beta}\cap\Delta$ looks like.

For $(5)$, for each $\alpha\in\Bbb R$ let

$$U_\alpha=\{\langle x,y\rangle:y\ge\alpha\}\qquad\text{and}\qquad R_\alpha=\{\langle x,y\rangle:x\ge\alpha\}\;,$$

and show that both sets are clopen (i.e., both closed and open). You can use them to show that if $p,q\in\Bbb R^2$, and $p\ne q$, the space can be partitioned into two clopen sets, one containing $p$ and the other containing $q$, thereby showing that $p$ and $q$ are not in the same connected component of the space.

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  • $\begingroup$ Thank you very much, Brian, this is an amazing answer! I was wondering, how did you come up with the clopem sets on part $(5)$? I guess there is no general strategy, but what should I look for when I need to find such sets? (In case I encounter problems like this in the future). $\endgroup$ – user313212 Aug 20 '16 at 20:57
  • $\begingroup$ @user313212: You’re very welcome. The topology is a bit reminiscent of the Sorgenfrey topology, which I know well; the sets $U_\alpha$ and $R_\alpha$ work for that, and it was just a moment’s work to check that they work equally well here. (I didn’t consciously make that connection at the time, but I’m reasonably sure that that background played a part in suggesting that I look at those sets.) Having already worked out the $\operatorname{cl}J$ for $J\in\mathfrak{B}$ helped. $\endgroup$ – Brian M. Scott Aug 20 '16 at 21:02
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To compare $\tau$ with the usual (Euclidean) topology $E$ on $R^2,$ observe that the set of open rectangles of the form $E(a,b,\alpha ,\beta)=(a,a+\alpha)\times (b,b+\beta),$ for $a,b\in R$ and $\alpha , \beta \in R^+$ , is a base for $E.$

Now $ E(a,b,\alpha ,\beta)=\cup \{J(a',b',\alpha -(a'-a), \beta- (b'-b)): a'\in (a,a+\alpha) \land b'\in (b,b+\beta)\}\in \tau.$ So $E\subset \tau.$

So $\tau$ is Hausdorff, because $E$ is Haudorff and $E\subset \tau.$

Also $E \ne \tau$ because $J(a,b,\alpha , \beta )\ \in \tau$ \ $E,$ because any $E$-nbhd of $(a,b)$ contains points $(x,y)$ with $x<a$ and $y <b.$

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