Show that $A$ is closed w.r.t. the upper limit topology

Let , $\tau_1$ be usual topology in $\mathbb R$ and $\tau_2$ be lower limit topology in $\mathbb R$. Let , $$A=\{x\in \mathbb R:4\sin^2x\le 1\}\cup \{\frac{\pi}{2}\}.$$ Then , (1) $A$ is closed w.r.t. $\tau_1$.

(2) $A$ is closed w.r.t. $\tau_2$.

Attempt :

The set $A=[-\pi/6,\pi/6]\cup \{\pi/2\}$..so it is closed w.r.t. $\tau_1$.

Now $A^c=(-\infty,-\pi/6)\cup(\pi/6,\pi/2)\cup(\pi/2,\infty)$. Now , $(-\infty,-\pi/6)$ is open in $\tau_2$ , but how the other two are openin $\tau_2$ so that we can say that $A^c$ is open in $\tau_2$ ?

• What does a basic open set look like in the lower limit topology? – Daniel Nov 2 '15 at 20:18

I don't quite agree with your $A$. Recall that $\sin(x)$ is periodic...

But as $f(x) = 4\sin^2(x)$ is continuous from the reals in $\tau_1$ to itself, and $C = (-\infty,1]$ is closed in the reals, $f^{-1}[C] = \{x \in \mathbb{R}: 4\sin^2(x) \le 1 \}$ is closed, and singletons are closed as well, so $A$, as a union of these, is closed in $\tau_1$. This follows without even computing exactly what $A$ is.

And as $\tau_1 \subset \tau_2$ ($(x,y) = \cup_{z \in (z,y)} [z,y)$), we know that any set open in $\tau_1$ will be open in $\tau_2$ and the same for closed as well.

• I can't understand that how $(x,y)$ can be expressed as the union of $[x,y)$ ? For ex. $(0,1)=$ ? – Empty Nov 3 '15 at 2:27
• $(0,1)=\cup_{a \in (0,1)} [a,1)$, check the two inclusions. – Henno Brandsma Nov 3 '15 at 7:03
• Ok..Am I wrong about closedness of $A$ w.r.t. $\tau_1$ ? – Empty Nov 3 '15 at 13:08
• @S.Panja-1729 I edited the answer for more about $A$. – Henno Brandsma Nov 3 '15 at 16:28
• Yes, you are. e.g.recall that $\sin(2\pi) = \sin(4\pi) = \ldots = 0$ so $2\pi, 4\pi,6\pi$ etc. are all in $A$. – Henno Brandsma Nov 3 '15 at 18:26