I need a check about this exercise:

Let $\mathcal{B}$=$\{[a,+\infty):a \in \mathbb{R}\}$ the basis that generate the topology $\tau$.

(i) Is $\tau$ finer or coarser than $\tau_e$?

(ii)Is $(-3,+\infty)$ open in $(\mathbb{R},\tau)$?

(iii) Find the interior of $(1,+\infty),(-\infty,0)$ in $(\mathbb{R},\tau)$

(iv) Find the closure of $\{1\}$,$[0,+\infty)$,$(-1,1)$ in $(\mathbb{R},\tau)$

(v)Say if $f:(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau), x \mapsto cx, c\in \mathbb{R}$ is an homeomorphism.

Here's my proof.

(i)Since $\tau_e \subset \tau$, $\tau$ is finer than the euclidian topology.

(ii)A set $\mathcal{A}$ is open iff for every $x$ in $\mathcal{A}$ there exists an open set $U$ such that $x \in U \subset \mathcal{A}$.

I think that $(-3,+\infty)= \bigcup_{n\geq1} [3-1/n,+\infty)$ and the union of open sets is an open set.



(iv) A closed set is like $(-\infty,a)$, $(-\infty,a]$, since it's complementar is open in $\tau$.




(v) If $c=0$ the function is the costant map $0$ and it's a banal homeomorphism.

Let's suppose $c\neq0$.

I have to show that $f$ is continuous and bijective and $f^{-1}$ is also continuos.

If $c>0$, $f^{-1}[a,+\infty)=[a/c,+\infty) \in \tau$, and $f^{-1}:(\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$, with $x\mapsto x/c$ is also continuos.

If $c<0$, for $a \in \mathbb R$, I got that $f^{-1}[a,+\infty)=(-\infty,a/c] \notin \tau$.

So $f$ is an homeomorphism iff $c \geq 0$.

  • $\begingroup$ you need to prove (i)... why is $\tau_e\subset \tau$? And if $\tau_e\subset \tau$ then $\operatorname{int}(-\infty,0)=(-\infty,0)$. $\endgroup$ – Masacroso Aug 21 '17 at 16:45
  • $\begingroup$ Are the sets in $\mathcal{B}$ supposed to be closed in the topology generated by $\mathcal{B}$? $\endgroup$ – Michael Lee Aug 21 '17 at 17:11
  • $\begingroup$ What is $\tau_e$? $\endgroup$ – 54321user Aug 21 '17 at 17:23
  • $\begingroup$ What is a "banal homeomorphism"? $\endgroup$ – drhab Aug 21 '17 at 17:27
  • $\begingroup$ @54321user the euclidean topology $\endgroup$ – Masacroso Aug 21 '17 at 17:35

It is mostly correct. A few remarks:

  1. While proving that (i) holds, you should have done something that you only do while proving that (ii) holds, namely that sets of the form $(a,+\infty)$ belong to $\tau$.
  2. Your justification of (ii) is clumsy. You state that a set is open if and only if it satisfies a certain condition and then you don't use that. And what happens is that $(-3,+\infty)=\bigcup_{n\in\mathbb N}\left[-3+\frac1n,+\infty\right)$.
  3. Yes, all sets of the form $(-\infty,a)$ and $(-\infty,a]$ are closed, but you should have added that, besides $\emptyset$ and $\mathbb R$, these are the only closed sets.
  4. If $c=0$, $f$ is not a homeomorphism (banal or othersise).
  • $\begingroup$ @feddy If this answer is satisfying for you then don't forget to accept it. Uptil now you have asked $10$ questions, but only one of them was accepted. If that's in accordance with your standards then they might be too high. $\endgroup$ – drhab Aug 21 '17 at 17:47
  • $\begingroup$ @JosèCarlosSantos: are the topologies not comparable as state drhab? I thought that every open set in the euclidian topology is in an open set like $[a,+\infty)$... nut now I'm no so sure $\endgroup$ – VoB Aug 22 '17 at 7:29
  • $\begingroup$ @feddy No, they are not comparable, since $[0,+\infty)\in\tau\setminus\tau_e$ and $(0,1)\in\tau_e\setminus\tau$. $\endgroup$ – José Carlos Santos Aug 22 '17 at 7:38
  • $\begingroup$ Clear, thanks! ;) $\endgroup$ – VoB Aug 22 '17 at 9:59

I preassume that $\tau_e$ denotes the common topology on $\mathbb R$.

(i) The topologies are not comparable since $[0,\infty)\in\tau$, $[0,\infty)\notin\tau_e$, $(0,1)\notin\tau$ and $(0,1)\in\tau_e$.

(ii), (iii), (iv) correct.

(v) If $c=0$ then $f$ is not a homeomorphism, since it is not bijective. What you say about $c>0$ and $c<0$ is correct.


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