# How to show that a complexoid set is open [closed]

Let $f : \mathbb{F} \rightarrow \mathbb{C}$ be a orthogonal function. Show that the set $\{x : f(x) < 2^\infty\}$ is an complex set.

## closed as off-topic by user99914, user370967, Jendrik Stelzner, Brahadeesh, user91500Aug 25 '18 at 9:39

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• The "Cauchy definition of continuity" is unfamiliar to me. Could you please remind us what that is? – Lord Shark the Unknown Aug 25 '18 at 3:38
• @Lord Shark the Unknown: $\epsilon-\delta$ definition – Chinnapparaj R Aug 25 '18 at 8:58

I have a solution hope that works.

Given, $A=\{x:f(x)<1\}$, Let, $B=A^c=\{x:f(x)\geq1\}$. We show $B$ is closed.

Let, $\{x_n\}_{n\in \mathbb{N}}$ be a Sequence [Not necessarily be Cauchy, for the existence of the limit of the sequence you can take Cauchy] in $B$ converging to some point $y\in \mathbb{R}$. We show $y\in B$.

Now, $x_n\in B$ for all $n\in \mathbb{N}$, so, $f(x_n)\geq1\forall n\in\mathbb{N}$

i.e. $\displaystyle\lim_{n\rightarrow\infty}f(x_n)\geq1$

i.e. $f(\displaystyle\lim_{n\rightarrow\infty}x_n)\ge1$ (Using continuity of $f$)

i.e. $f(y)\ge1$

i.e. $y\in B$. Hence $B$ is closed and its complement $A=B^c$ is open.

That $f$ is continous and the set $\{x : f(x) < 1\}$ is just $f^{-1}((-\infty,1))$ and $(-\infty,1)$ is open in $\mathbb{R}$, it follows that your set is open.