1
$\begingroup$

In the real plane $\mathbb{R}^2$ is considered the topology $\tau$ whose basis consists in the open squares centered at $p$ with lengths $\varepsilon >0$, excluding the points on the two diagonals other than $p$.

$a)$ Study the continuity of $f: (\mathbb{R},\tau_u) \to (\mathbb{R}^2,\tau)$ defined by $f(t)=(t,\lambda t)$ for $\lambda \in \{0,1\}$.

$b)$ Show that $\tau$ is not a product topology.

$c)$ Does $(\mathbb{R}^2,\tau)$ verify the first countability axiom? Is it separable? Does it verify the second countability axiom?

For the part $a)$, I think that the function $f$ is indeed countinuous. Let $U$ be an open set in $(\mathbb{R}^2,\tau)$. Then $$f^{-1}(U)=\left\{ t \mid f(t)\in U \right\}=\left\{ t \mid (t,\lambda t) \in S \right\}$$ and that set is (if the square's center is $p=(a,b)$, equal to $(a-\varepsilon/2,a+\varepsilon/2)$ (or at least I think so).

For part $b)$, I don't have a clue, so an idea of how to prove it would be helpful.

For part $c)$, I think the open squares as in $\tau$ with the requisite that the epsilons are rational numbers might work as a local basis for any point of $\mathbb{R}^2$, and I don't see any reason why in the same way, the subset of the basis consisting in the open squres centered at $p$ with lengths being rational numbers wouldn't work as a countable basis for the topology. Hence, first and second countability axioms hold.

But that implies that $(\mathbb{R}^2,\tau)$ is separable, which makes me think that the second countability axioms does not hold and I made some mistake above. In any case, I believe that, as in $\mathbb{R}^2$ with the usual topology, $\mathbb{Q}^2$ is a countable dense subset.

Thanks in advance!

$\endgroup$
1
$\begingroup$

Note that part (a) actually asks you about two different functions: one, with $\lambda=0$, takes $t\in\Bbb R$ to $\langle t,0\rangle$ on the $x$-axis, and the other, with $\lambda=1$, takes $t$ to $\langle t,t\rangle$ on the diagonal line $y=x$. I’ll denote the first of these functions by $f_0$ and the second by $f_1$. Also, for each $p=\langle a,b\rangle\in\Bbb R^2$ and $\epsilon>0$ I’ll let

$$B(p,\epsilon)=\left\{\langle x,y\rangle\in\Bbb R^2:|x-a|,|y-b|<\frac{\epsilon}2\text{ and }|x-a|\ne|y-b|\right\}\;;$$

this is the open square centred at $p$ with sides of length $\epsilon$, excluding the two diagonals except for the point $p$ itself. Let $\mathscr{B}$ be the family of all of these basic open sets $B(p,\epsilon)$.

The function $f_1$ is not continuous: if $t$ is any real number, $B\big(f_1(t),1\big)$ is a basic open nbhd of $f_1(t)$, and you should check that

$$f^{-1}\left[B\big(f_1(t),1\big)\right]=\{t\}\;,$$

which is certainly not open in the usual topology on $\Bbb R$. (It may help to realize that $f_1[\Bbb R]$ is the diagonal, the line $y=x$, and that if $p$ is any point on this diagonal, then $B(p,r)=\{p\}$ for every $r>0$.

The function $f_0$, on the other hand, is continuous. Let $B=B(p,r)$ be a basic open set in $\Bbb R^2$, where $p=\langle a,b\rangle$; we want to show that $f_0^{-1}[B]$ is open in $\Bbb R$. This isn’t hard, but we do have to consider several cases.

  • If $|b|\ge r$, you can easily check that $f_0^{-1}[B]=\varnothing$, since in that case $B\cap f_0[\Bbb R]=\varnothing$: $f_0[\Bbb R]$ is the $x$-axis, and in this case $B$ does not intersect the $x$-axis.

  • If $b=0$, you can easily check that $f_0^{-1}[B]=\left(a-\frac{r}2,a+\frac{r}2\right)$.

  • The trickiest case is the one in which $0<|b|<r$. I’ll leave it to you to verify that in this case $$f_0^{-1}[B]=\left(a-\frac{r}2,a-|b|\right)\cup\left(a-|b|,a+|b|\right)\cup\left(a+|b|,a+\frac{r}2\right)\;;$$ I recommend drawing some sketches to help.

As you can see, in each case $f_0^{-1}[B]$ is open in the usual topology on $\Bbb R$. If $U\in\tau$, then $U$ is a union of members of $\mathscr{B}$, so $f_0^{-1}[U]$ is a union of sets of the form $f_0^{-1}[B]$ with $B\in\mathscr{B}$; we’ve just seen that all of these are open in $\Bbb R$, so $f_0^{-1}[U]$ is open in $\Bbb R$ as well, and $f_0$ is continuous. In fact, $f_0$ is a homeomorphism from $\langle\Bbb R,\tau_u\rangle$ to the $x$-axis with the relative topology that it inherits from $\langle\Bbb R^2,\tau\rangle$: it’s clearly a bijection to the $x$-axis, and it’s not hard to check that its inverse is continuous as a function from the $x$-axis in $\langle\Bbb R^2,\tau\rangle$ to $\langle\Bbb R,\tau_u\rangle$.

Note that in your attempt you started with an arbitrary $U\in\tau$ and assumed that it was actually one of the basic open sets in $\mathscr{B}$. This isn’t true: it must be a union of members of $\mathscr{B}$, but it need not be in $\mathscr{B}$ itself. (After all, it could be all of $\Bbb R^2$, for instance.) However, the argument that I sketched in the previous paragraph works in general to show that if the inverse images of all basic open sets are open, then the inverse images of all open sets are open, and the function is continuous. It’s only ever necessary to check the inverse images of some base for the topology of the codomain.

Let’s deal with (c) next. You are correct in thinking that $\Bbb Q^2$ is dense in $\Bbb R^2$ with respect to the topology $\tau$ and hence that $\langle\Bbb R^2,\tau\rangle$ is separable. One way to see this is to observe that each $\tau$-basic open set $B(\langle a,b\rangle,r)$ contains the open triangle with vertices $\left\langle a-\frac{r}2,b+\frac{r}2\right\rangle$, $\left\langle a+\frac{r}2,b+\frac{r}2\right\rangle$, and $\langle a,b\rangle$, which is open in the usual topology and therefore contains points of $\Bbb Q^2$.

It takes some work to carry out the details, but it can be shown that for each $p\in\Bbb R^2$

$$\left\{B\left(p,\frac1n\right):n\in\Bbb Z^+\right\}$$

is a local base at $p$ and hence that $\langle\Bbb R^2,\tau\rangle$ is first countable. You have to show that if $B$ is any member of $\mathscr{B}$ that contains $p$, including those not centred at $p$, then there is an $n\in\Bbb Z^+$ such that

$$B\left(p,\frac1n\right)\subseteq B\;.$$

The case in which $B$ is not centred at $p$ actually turns out to be quite straightforward: in that case $p$ is not on either diagonal of $B$, so there is even an $n\in\Bbb Z^+$ such that the Euclidean open square of side $\frac1n$ centred at $p$ is contained in $B$.

However, $\langle\Bbb R^2,\tau\rangle$ is not second countable. One easy way to show this is to let $D=\{\langle x,x\rangle:x\in\Bbb R\}$, the diagonal that is the graph of $y=x$, and observe that if $p\in D$ and $r>0$, then $B(p,r)\cap D=\{p\}$. (I made this observation before in connection with the function $f_1$.) Suppose that $\mathscr{U}$ is base for $\tau$. Then for each $p\in D$ there must be a $U_p\in\mathscr{U}$ such that $p\in U_p\subseteq B(p,1)$. I leave it to you to verify that if $p,q\in D$ and $p\ne q$, then $U_p\ne U_q$, and therefore $|\mathscr{U}|\ge|D|=|\Bbb R|$, so that $\mathscr{U}$ must be uncountable. This shows that the space has no countable base.

Finally, we’ll show that $\tau$ cannot be a product topology. Suppose that there are topologies $\tau_0$ and $\tau_1$ on $\Bbb R$ such that $\langle\Bbb R^2,\tau\rangle$ is the product of $\langle\Bbb R,\tau_0\rangle$ and $\langle\Bbb R,\tau_1\rangle$. For each $t\in\Bbb R$ let $V_t=\{t\}\times\Bbb R$ and $H_t=\Bbb R\times\{t\}$, with the relative topologies that they inherit from $\langle\Bbb R^2,\tau\rangle$. It’s a basic fact about product topologies that each $V_t$ is homeomorphic to $\langle\Bbb R,\tau_1\rangle$ and each $H_t$ to $\langle\Bbb R,\tau_0\rangle$. However, it’s not hard to show that in fact each $V_t$ and $H_t$ is homeomorphic to $\langle\Bbb R,\tau_u\rangle$; we did most of this in the case of $H_0$ in proving above that $f_0$ is continuous.

This means that $\tau_0=\tau_1=\tau_u$. But then the topology on the product of $\langle\Bbb R,\tau_0\rangle$ and $\langle\Bbb R,\tau_1\rangle$ is just the usual topology on $\Bbb R^2$, which certainly is not $\tau$. Thus, $\tau$ cannot be a product topology.

$\endgroup$
  • $\begingroup$ Your proof that the weight is $|\mathbb R|$ is reminiscent ot the method of showing that the weight of the Sorgenfrey line is $|\mathbb R|$. $\endgroup$ – DanielWainfleet Sep 13 '16 at 6:07
  • 1
    $\begingroup$ @user254665: It’s even more like the easy way to show that the Sorgenfrey plane has weight $2^\omega$, using the points on the line $y=-x$: in both cases you have a closed, discrete set of cardinality $2^\omega$. $\endgroup$ – Brian M. Scott Sep 13 '16 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.