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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.

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  • $\begingroup$ Note that $(1/2, \infty) \notin \tau$, since $0.6 \in (1/2, \infty)$ but $0.6^2 = 0.36 \notin (1/2, \infty)$. Likewise, $\mathbb R \setminus \{1\} \notin \tau$ because it contains $-1$ but not $(-1)^2 = 1$. $\endgroup$ – Adriano Aug 21 '17 at 8:09
  • $\begingroup$ You're right, thank you. So Int($(1/2,+\infty)=(1,+\infty)$, right? But what about the closure of the singleton {1}? I don't know what to do $\endgroup$ – VoB Aug 21 '17 at 8:16
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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.

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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 $U$.

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  • $\begingroup$ Hi, for (ii) I think that $Int(1/2,+\infty)=(1,+\infty)$. (iii) Sincerly, I don't know ho to move (iv) How can I show that this works for every open set $U$? $\endgroup$ – VoB Aug 21 '17 at 8:21

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