I know the following type of questions has been asked many times on Math SE. Please let me apologies if my questions seems to be of a repeating nature. For the purpose of my question, we won't consider metric spaces at all.

The notion of open and closed sets between general topological spaces are just called open sets in the three criteria within the definition of a topological spaces.
Often time, when we are asked to prove a set is open, we can try to prove the set's complement is closed, and hence take the set's complement. In vice versa manner, similar strategy can be used for the case of a closed set. Also, we are told, an open set is not the opposite of a closed set. So a set that is not open is not the same as saying that a set is closed. Here are my points of confusions or questions.
There are a collection of results stating necessary and sufficient conditions for when one can consider sets to be either open or closed, and they are stated involving notions of interior, closure, limit point and boundary of the set. This also applies to both cases of a function being an open or a closed map.

Below, I have listed an example of a theorem respectively for the cases of open sets, closed sets, and open and closed map. I have stated a standard theorem for each case, and in the case of say a set is not open, from the theorem stated, does how I change the phrasing reflect the consequences of how the theorem would not hold. Basically, what I am interested in is, to say a set B is not open, how does it affect the collection of results stating when a set can be consider to be an open set mathematically speaking, and also with respect to the relations to each other between set B's boundary, closure, interior and limit points. So in the case of set B being not open, does it mean it is not a subset of its interior and as a result, it contain some of its boundary. Similarly for a closed set, what else can we say about it. I ask this because if in the case of trying to prove that a set is open, but you are having difficulty proving its complement is closed. Well, what about assuming that is is not open, then what can be stated about a set that is known only to be not open and nothing else.

A set $B$ is not open means $B\subsetneqq \mathring{B}$

A set $C$ is closed iff $C=\overline{C}$

A set $B$ is not closed means $\overline{B}\subsetneqq C$

Also for the cases of both an open and a closed maps; we define a function to be an open map as:

Let $(X,\mathcal{T})$ and $(Y,\mathcal{U})$ be topological spaces, A function $f:X\rightarrow Y$ is open provided if $U \in \mathcal{T}$ then $f(U)\in\mathcal{U}$

hence when we talk about $f$ is not an open map, does it mean either one of the following:

  1. A function $f:X\rightarrow Y$ is not open provided if $U \in \mathcal{T}$ is an open subset in $X$, then $f(U)\notin\mathcal{U}$, meaning $f(U) \subsetneq \mathring{f(U)}$
    a) $f(U)$ is not open means: if $U \in \mathcal{T}$ is an open subset in $X$, then $f(U) \subsetneqq \mathring{f(U)}$
    b) if $U \notin \mathcal{T}$, which means $U \subsetneq \mathring{U},$ and $f(U) \subsetneqq \mathring{f(U)}$

Similarly, we define a function to be a closed map to mean:

  1. $C$ is a closed subset of $(X,\mathcal{T})$, then $f(C)$ is a closed subset of $(Y,\mathcal{U})$, meaning $f(C)=\overline{f(C)}$
    So to speak of a function being not a closed map, does it mean either:
    a) A function $f:X\rightarrow Y$ is not closed provided if $C$ is a closed subset of $(X,\mathcal{T})$, and $f(C)$ is not a closed subset of $(Y,\mathcal{U})$, meaning $\overline{f(C)} \subsetneqq f(C)$
    b) A function $f:X\rightarrow Y$ is not closed provided if $C$ is not a closed subset of $(X,\mathcal{T})$, $\overline{C} \subsetneqq C$, and $f(C)$ is not a closed subset of $(Y,\mathcal{U})$, meaning $\overline{f(C)} \subsetneqq f(C)$

Thank you in advance

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    $\begingroup$ @Daniele Tampieri thank you for the edit. I hope i got all the format and markup code correctly. I am know certain things would work for a regular LaTex compiler but is not needed for typing math here on Math SE. I just hope I don't come across too many of those every time I ask a math question on here.. $\endgroup$
    – Seth Mai
    Sep 11 '19 at 9:52
  • $\begingroup$ Seth, your edit technique is OK, including all the markup codes you choose to use: I only added a few ones in order to format your question the way I though you desired. Do not worry about it: in general there are many tricks that you'll notice by editing Q&A here. I am still in the process of learning them (and this will probably continue for a long time...). Have a nice day. $\endgroup$ Sep 11 '19 at 10:03
  • $\begingroup$ @Daniele Tampieri thank you again. :) $\endgroup$
    – Seth Mai
    Sep 11 '19 at 10:22

I'm not sure I understand your question correctly, but let me give a try on answering anway.

Basically, given an arbitrary set $S$ in a topological space, you can partition the space into three sets:

  • The interior, which is the largest open subset of $S$. Here “largest” refers to the subset partial order.

  • The exterior, which is the largest open set disjoint with $S$.

  • The boundary, which is the set of all points neither in the interior nor in the exterior.

Now from the definition, you can immediately see several things:

  • The interior and exterior are by definition open, while the boundary is closed (it is the complement of an open set, the union of interior and exterior).

  • $S$ is the union of its interior and a subset of its boundary.

  • $S$ is open if and only if it is disjoint with its boundary. $S$ is closed if and only if it is a superset of its boundary. It is both open and closed at the same time if and only if the boundary is empty.

  • Since the interior of $S$ is by definition a subset of $S$, if $S$ is a subset of its interior, it must be equal to its interior. And that obviously means it is open.

  • Moreover, the closure of $S$ is the smallest closed superset of $S$, which means its complement is exactly the exterior. Thus the closure is the union of interior and boundary, or equivalently, the union of the set with its boundary (since any non-interior points of the set are at its boundary anyway).

  • Therefore a set is always a superset of its interior and a subset of its closure: $$S^\circ \subseteq S \subseteq \overline S$$

Now on your theorems:

  • Since $B^\circ\subseteq B$, the relation $B\subsetneq B^\circ$ is always false. The condition you want is $B^\circ\subsetneq B$.

  • It is true that $C$ is closed iff $C=\overline C$.

  • I assume the final $C$ should have been $B$ again. Then, as before, the relation you wrote is always false; you want $B\subsetneq \overline B$ instead.

In the following I'll assume the corrected versions throughout.

On the open/closed functions:

  1. Mind the quantors: A function is open if for all open sets the image is open. Thus a function is not open if there exists an open set for which the image is not open. In particular, there may still be open sets whose image is open.

    For example, the real function $x\mapsto x^2$ is not open because the open interval $(-1,1)$ gets mapped on the interval $[0,1)$ which is not open. Nevertheless the interval $(1,2)$ gets mapped to the interval $(2,4)$ which is open.

    Indeed, the latter is an explicit counterexample to your claim a)

    For a counterexample for claim b), take the same function, and use the set $X=((-2,-1)\cap\mathbb Q)\cup((1,2)\setminus\mathbb Q)$, that is, $X$ consists of all rational numbers in $(-2,-1)$ and all irrational numbers in $(1,2)$. This set is clearly not open in $\mathbb R$, however its image under the square function is the set $(1,4)$ which is open.

  2. Here you've formulated more ambiguously, so a) could be read as the correct statement (“there exists a closed set $C$ such that …”), or as containing the same error as you made in the open functions (“if $C$ is closed, then …” or equivalently “for all closed sets $C$, …”).

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    $\begingroup$ $B$ is not open is equivalent to $B \nsubseteq B^\circ$: there is a non-interior point, just as $B$ is not closed is equivalent to $\overline{B} \nsubseteq B$ as well, contrary to your claim.. $\endgroup$ Sep 11 '19 at 22:34
  • $\begingroup$ @HennoBrandsma Henno, thank you for the reply. I just would to clarify, so for a set to not being open, it still always contains its own interior, but it is not a subset of its own interior. Likewise for a set that is closed, the closure of the set always the set itself, but the set itself is a superset of its own closure. Also, would the following statements be true as well, If a set is not open, then its own boundary is not a subset of the set's complement... $\endgroup$
    – Seth Mai
    Sep 12 '19 at 3:01
  • $\begingroup$ @HennoBrandsma Also would the following be correct as well: If a set is closed then, it does not contain its own boundary. If a set is neither open nor closed, then its own boundary is not empty. Lastly, for a function to be either an open map nor a closed map, how would that be phrased precisely in set builder notation. $\endgroup$
    – Seth Mai
    Sep 12 '19 at 3:04
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    $\begingroup$ @SethMai A set is always a subset of its closure, but a non-closed set has a point in the closure that's not yet in the set, so $\overline{A} \nsubseteq A$. A set is closed iff $\partial A \subseteq A$ so a set is not closed iff there is a point in $\partial A$ not in $A$, so $\partial A \nsubseteq A$ etc. $\endgroup$ Sep 12 '19 at 6:58
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    $\begingroup$ @SethMai as the answer said, $f:(X,\tau_X) \to (Y, \tau_Y)$ is not open iff $\exists O \in \tau_X: f[O] \notin \tau_Y$ and similarly for closed. There needs to be at least one counterexample for a universal statement to fail. $\endgroup$ Sep 12 '19 at 7:01

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