General topology exercise (closure,interior, continuity) I'd like a check about this exercise

Let $\tau=$ {$U \subset \mathbb{R}: U \in \tau_e, \forall x \in U, x^2 \in U$}, with $\tau_e$ the standard euclidian topology.

(i)Is $\tau$ finer or coarser than $\tau_e$?
(ii)Find $Int[-1,1]$, $(-\infty,1)$, $Int(1/2,+\infty)$, $Int[-2,-1]$.
(iii)Find {$\overline{x}$}
(iv) Show that $f:(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$, $f(x)=x^s$, $s \in \mathbb{N}$ is continuous.

Here's my solution:
By the definition of this topology, the open sets are $\mathbb{R}$, $\emptyset$,and every subset of $(-1,1)$. For example, $(0,1) \in \tau$
(i) Since $\tau \subset \tau_e$, $\tau_e$ is finer than $\tau$. In fact, every open set in $\tau$ is open also in $\tau_E$.
(ii)
$Int[-1,1]=(-1,1)$.
$Int(-\infty,1)=(-1,1)$
$Int(1/2,+\infty)=(1/2,+\infty)$ 
$Int[-2,-1]=\emptyset$. In fact, if $A \subset \tau$, $A\subset [-2,-1]$, then {$4$} $\in A$, which is a contadiction.
(iii) I say the closure of the singleton {$1$} is {$1$}, because {$1$}$^c=\mathbb{R} \setminus$ {$1$}, which is open in $\tau$.
(iv)
For every $s \in \mathbb{N}$, the pre-image of $\mathbb{R}, \emptyset$ are respectively $\mathbb{R}, \emptyset$.
If I take $U =(-1,1)$, $f^{-1}(U)=(-1,1) \in \tau$. And if $U=(0,1), f^{-1}(U)=(0,1)$, for all natural $s$.
So $f$ is continuous.
 A: 
By the definition of this topology, the open sets are $\mathbb{R}$, $\emptyset$,and every subset of $(-1,1)$

False. The set $A=(\frac13,1)$ is not in $\tau$, since $x=\frac12\in A$, but $x^2=\frac14\notin A$.
On the other hand, the set $(0,\infty)$ is in $\tau$, but it does not match any of your descriptions.


(i) Since $\tau \subset \tau_e$, $\tau_e$ is finer than $\tau$. In fact, every open set in $\tau$ is open also in $\tau_E$.

Correct.

All your answers for (ii) are based on your first falacy, so you also need to rethink them.


(iii) I say the closure of the singleton {$1$} is {$1$}, because {$1$}$^c=\mathbb{R} \setminus$ {$1$}, which is open in $\tau$.

Incorrect. The set $\mathbb R\setminus\{1\}$ is not open in $\tau$, because $x=-1\in\mathbb R\setminus\{1\}$, but $x^2=1\notin \mathbb R\setminus \{1\}$.


(iv)
  For every $s \in \mathbb{N}$, the pre-image of $\mathbb{R}, \emptyset$ are respectively $\mathbb{R}, \emptyset$.
If I take $U =(-1,1)$, $f^{-1}(U)=(-1,1) \in \tau$. And if $U=(0,1), f^{-1}(U)=(0,1)$, for all natural $s$.
So $f$ is continuous.

Inorrect proof. You did not show that $f^{-1}(U)$ is open for every open set $U$.
A: Alternative for (iv):
Let $g:\mathbb R\to\mathbb R$ be prescribed by $x\mapsto x^2$.
Then  $\tau=\{U\in\tau_e\mid U\subseteq g^{-1}(U)\}$. 
To prove that function $f$ prescribed by $x\mapsto x^s$ is continuous it is enough to verify that for an arbitrary $U\in\tau$ we have $f^{-1}(U)\in\tau$. 
From $U\in\tau_e$ and the obvious fact that $f:\langle\mathbb R,\tau_e\rangle\to\langle\mathbb R,\tau_e\rangle$ is continuous we find that $f^{-1}(U)\in\tau_e$. 
What remains now is proving that: $$f^{-1}(U)\subseteq g^{-1}\left(f^{-1}(U)\right)=(f\circ g)^{-1}(U)$$ which can be done on base of $U\subseteq g^{-1}(U)$.
Here $f\circ g$ is prescribed by $x\mapsto x^{2s}$ so this means that actually on base of $x\in U\implies x^2\in U$ for every $x\in\mathbb R$ it must be shown that $x^s\in U\implies x^{2s}\in U$ for every $x\in\mathbb R$.
Well, that is quite obvious.
