# Standard topology exercise about continuous function and interior

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 $$-1 and $$1, I got the thesis.

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

$$\operatorname{Int}([-2,3]=(-2,3)$$.

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

(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.

• 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 $[-2,3] \notin \tau_e$
– VoB
Aug 20, 2017 at 13:42
• @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. Aug 20, 2017 at 13:46
• Understood ;). Anyway, I can't see why the interior of $[-2,3]$ is $[-2,3]$. This set contains $[-1,1]$, but $[-2,3] \notin \tau_e$
– VoB
Aug 20, 2017 at 13:48
• @feddy My mistake. I've edited my answer. Note that I wrote “Indeed”; I was agreeing with you. Aug 20, 2017 at 13:52
• Ok, it's all clear now. Thanks so much ;)
– VoB
Aug 20, 2017 at 13:53