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!
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 $\langle\Bbb R,\tau_u\rangle$.
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.