I'd like a check about this little exercise. (Sorry for my bad english, I'm an italian student)

Let $\tau=\{U \in \tau_e: [-1,1] \subset U\} \cup \{ \emptyset\}$, with $\tau_e$ the standard euclidian topology.

(i) Say if $\tau$ is finer than $\tau_e$.

(ii)Show that for all $U \in \tau$, with $U \neq \emptyset$, there exists $a,b \in \mathbb{R}$ such that $[-1,1] \subset (a,b) \subset U$.

(iii) Find $\operatorname{Int}(0,1)$, $\operatorname{Int}[-2,3]$, $\operatorname{Int}([-1,1] \cup [5,7])$

(iv)Determine the closure of $\{x\}$ in $\tau$, with $x \in \mathbb{R}$

(v) Is the function $f: (\mathbb{R},\tau) \rightarrow (\mathbb{R},\tau)$ such that $x \mapsto x+2$ continuous? And $x \mapsto x^2$?

Here's my solution

(i) I have that $\tau \subset\tau_e$, so $\tau_e$ is finer than $\tau$. In fact, for all $B \in \tau$, and for all $x \in B$, there exists $ C \in \tau_e$ such that $x \in C \subset B$

(ii) Let $U \in \tau$, $U:=(c,d)$. Since $U$ is an open set in the euclidian topology, I can always found an open set $(a,b) \subset U$. Now, if I take $-1<a<c$ and $1<b<d$, I got the thesis.

(iii) $\operatorname{Int}(0,1)=\emptyset$, because every open set in $\tau$ must contain $[-1,1]$


$\operatorname{Int}([-1,1] \cup [5,7]) = \emptyset$

(iv) If $x> 1$, then $\overline{\{x \}}=[x,+\infty)$

If $x<-1$, then $\overline{\{x \}}=(-\infty,x]$

If $x \in [-1,1]$, then its closure is $\mathbb{R}$.

(v) The first function is not continuous because the pre-image $f^{-1}(-3/2,3/2)$ doesn't contain $[-1,1]$.

But for $f(x)=x^2$, and $[-1,1] \in (a,b)$, I got that $f^{-1}(a,b)=(-\sqrt(a), \sqrt(a))$, and for $a>1$ this set contains $[-1,1]$, so $f$ is continuous. Also using unlimited open sets such as $(-\infty,a)$, $(b,+\infty)$ that contains $[-1,1]$, the pre-image is always an open set. So $f(x)=x^2$ is continuous.


(i) Right, but your second sentence is irrelevant.

(ii) Wrong. First of all, you cannot say that your arbitrary open set $U$ has the form $(a,b)$; not all open sets are like that.

Let $U$ be an onpen set that contains $[-1,1]$. For each $x\in[-1,1]$, there is a $\varepsilon_x>0$ such that $(x-\varepsilon_x,x+\varepsilon_x)\subset U$. Then $\bigcup_{x\in[-1,1]}(x-\varepsilon_x,x+\varepsilon_x)$ is an open interval containing $[-1,1]$.

(iii) Every non-empty open set must contain $[-1,1]$; otherwise, your proof that $(0,1)^\circ=\emptyset$ is correct.

Indeed, $[-2,3]^\circ=(-2,3)$ and $\bigl([-1,1]\cup[5,7]\bigr)^\circ=\emptyset$.

(iv) If $x\notin[-1,1]$, $\{x\}$ is closed, and therefore $\overline{\{x\}}=\{x\}$. Otherwise, yes, $\overline{\{x\}}=\mathbb R$.

(v) Yes, the first function is not continuous because the pre-image of $\bigl(-\frac32,\frac32\bigr)$ is not an open set.

The second function is indeed continuous.

  • $\begingroup$ Thanks for the answer. I have a doubt about (iii). You say that every non-empty open set must contain $[-1,1]$. But this non-empty set must be in $\tau_e$. I mean that $[-2,3] \notin \tau_e$ $\endgroup$ – VoB Aug 20 '17 at 13:42
  • $\begingroup$ @feddy You wrote that “every open set in $\tau$ must contain $[−1,1]$”. I just noted that the empty set (which is an element of $\tau$, of course) is an exception. $\endgroup$ – José Carlos Santos Aug 20 '17 at 13:46
  • $\begingroup$ Understood ;). Anyway, I can't see why the interior of $[-2,3]$ is $[-2,3]$. This set contains $[-1,1]$, but $[-2,3] \notin \tau_e$ $\endgroup$ – VoB Aug 20 '17 at 13:48
  • $\begingroup$ @feddy My mistake. I've edited my answer. Note that I wrote “Indeed”; I was agreeing with you. $\endgroup$ – José Carlos Santos Aug 20 '17 at 13:52
  • $\begingroup$ Ok, it's all clear now. Thanks so much ;) $\endgroup$ – VoB Aug 20 '17 at 13:53

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