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If X is space and A is a non-empty proper open subset that is both open and closed, then is the set B = X - A both closed and open simply because the definition of a closed set is that its complement is open?

I understand it if I just follow the definition, but intuitively it doesn't make much sense, since it doesn't mention the openness/closeness of X and I don't see how the closed sets fit into the definition of a topology.

This question actually came up when I read the definition of a connected space, that is:

definition of connected space

I can use the line of reason above to understand the definition of a connected space (i.e. the complement of B is A, which is both open and closed, hence B is both closed and open, hence A and B constitute a separation of X and X is not connected), but other than that it doesn't provide much more information.

Also, I'm new to topology, so everytime I need to determine whether a certain set is closed or open, I have the habit of thinking of the "Venn diagram", and the notion of an open set in $\Re^2$ (i.e. whether I can draw a ball around every the point of the set that fits inside the set). Clearly this doesn't apply in lots of cases. Is there a better way to think about these concepts?

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  • $\begingroup$ Useful question. +1 $\endgroup$ – user370967 Jan 25 '18 at 22:04
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    $\begingroup$ In this context $X$ is a topological space, which is always both open and closed in itself just by the definition of a topology on $X$. $\endgroup$ – Jair Taylor Jan 25 '18 at 22:05
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    $\begingroup$ As you mentioned, a proper closed and open subset creates a separation on $X$, which means that $X$ is a union of two disjoint open sets. For me this is an intuitive way to think of connectness or in this case disconnectness $\endgroup$ – MPos Jan 25 '18 at 22:09
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One way to understand connected spaces better is the "Main Theorem on Connected Space", which says:

Let $X$ and $Y$ be topological spaces, and $f:X\to Y$ a continuous function. If $X$ is connected, then $f(X)$ is connected.

This is a generalization of the Intermediate Value Theorem (just let $X,Y$ be $\mathbb{R}$ with the euclidean topology). So, connected spaces are just "where the IVT holds".

This reasoning can even help when looking at disconnected spaces. The Subspace Topology is defined as:

Let $(X,\tau)$ be a top. space, and $S\subseteq X$ any subset (not necessarily open). Then the subspace topology is: $$\tau_S = \{S\cap U\mid U\in\tau\}$$

Now, if $S$ is chosen to be open, it turns out that $\tau_S$ is just all of the open sets in $\tau$ that are subsets of $S$. The subspace topology has one property I'll single out for this:

If $f:X\to Y$ is continuous, then the restriction to $S$ is continuous (with respect to the subspace topology, and $S$ need not be open in the "normal" topology on $X$).

So:

  1. Given a connected space $X$, we "have the IVT" already by the Main Theorem on Connected Spaces.

  2. Given a disconnected space $X$, we have that for any non-trivial clopen set $S$ (with no proper non-trivial clopen subsets) that $f\mid_S$ "satisfies the IVT", where $f\mid_S:(S,\tau_S)\to Y$. This is by applying the quoted result on subspaces, noticing that this space is now connected, and then applying the main theorem on connectedness.

Of course, this doesn't help with all disconnected spaces (consider the discrete topology on any set, then we get that "the IVT holds if we restrict functions to a single point domain", which is quite disinteresting.

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Surely, you can imagine open sets from $ℝ^n$ – nice domains containing nothing from their boundary like open ball or a cube without boundary. Similarly, closed sets contain everything from their boundary – closed ball, cube, a circle. But the notion of a topological space is an abstraction – it abstracts the main propertied of open/closed sets from metric spaces. A set is open just if it is in the collection of all open sets – the topology – and this may be any collection of sets satisfying the axioms. And closed sets are their complements – they form a collection satisfying the dual axioms.

All four combinations are possible – there are open sets that are not closed, there are closed sets that are not open (and these are essentially the only open and closed sets in a connected space, e.g. $ℝ^n$). Usually, most of the sets are neither open nor closed (even though there is a class of weird spaces defined by the condition that every subset is open or closed, they are caled door spaces). And finally, there are clopen sets – those that are both open and closed.

Surely, a complement of a clopen set is also clopen, and $∅$ and $X$ are always clopen. As @MPos comments, if you have a proper clopen set $A ⊆ X$ and you put $B := X \setminus A$, then $X$ is the disjoint union of the two clopen sets $A$ and $B$. The point is that the two parts are completely independent. You can even forget the topology of $X$ and remember just the subspace topologies on $A$ and $B$. The topology of $X$ can be reconstructed from these – $X$ is the topological sum of $A$ and $B$ (topological sum is one of four fundamental constructions on topological spaces). You can imagine it like this: the space $A$ is a clump in your left hand, $B$ is a clump in your right hand, and $X$ are just these two clumps together, one beside the other, with a gap between. A simple example is a space $(0, 1) ∪ (2, 3)$ – its topology comes from $ℝ$, but now it is a space on its own, having two (and hence clopen) components.

Connectedness means that this situation cannot happen – the given space cannot be decomposed into two nonempty clopen parts. It is just one clump and cannot be viewed as two separate clumps in any way. Also, one fundamental fact is that connectedness is preserved by continuous images – more details in @Mark's answer.

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