# Clopen Sets and Sets being Neither Open Nor Closed

I am very new to Topology. I do not have a college math degree, only highschool math. I have a real hard time understanding how a set can be both open and closed. And set being neither open nor closed.

Is it correct to think of a set being both open and closed as below?

Let's consider a topology over two sets (A and B) that are disjoint. For example, A = (1,2) and B = (4,5). Both the sets are open as they don't contain their limit points. However, by definition, a set is closed if that set's compliment is open. Therefore, A is closed because A's compliment (which is B) is open and B is closed because its compliment is open.

I am yet to come up with an example for sets that are neither open nor closed. It would be really great if someone can give a simple example for that.

• The rationals are neither open nor closed. Their interior is empty, their closure is all of $\mathbb{R}$. Dec 17, 2020 at 1:45
• The empty set and the entire set are both open and closed. And in this context, complement does not have an i Dec 17, 2020 at 1:45
• Compliment: “Your post is well crafted and you provide appropriate context and evidence of work. Well done!”. Complement: the stuff that is not part of that being considered. Dec 17, 2020 at 1:46
• @ArturoMagidin thanks. I was misreading the question. Dec 17, 2020 at 1:49
• When you talk about sets being open, closed, or neither, it's important to be clear about which topology you mean. To take your example, $A$ is clopen in the subspace topology induced by the usual topology on $\Bbb R$ on the subspace (of $\Bbb R$) $A \cup B$, but $A$ is merely open (not also closed) considered as a subspace of $\Bbb R$. Dec 17, 2020 at 2:13

In the real numbers $$\mathbb{R}$$, any half-open interval of the form $$[a,b)$$ or $$(a,b]$$ where $$a is neither open nor closed.
The definition of an open set in the real line is a set that is equal to the set of its inner points, that is: a set $$A$$ is open if for every $$a\in A$$ there exist a small number $$\varepsilon$$ such that $$(a-\varepsilon,a+\varepsilon)$$ is contained in $$A$$, a set is closed if its complement is open. Take $$A=[0,1)$$ in $$\mathbb R$$, this set is not open since $$(0-\varepsilon,0+\varepsilon)$$ will never be contained in $$A$$ but $$0\in A$$. And it's complement is not open since $$1\in A^c$$ but $$(1-\varepsilon,1+\varepsilon)$$ cant be contained in $$A^c$$, so $$[0,1)$$ is neither open or closed.