# On Connectedness, Intuitively

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Today's post is a bit of a ramble, but my goal is to uncover the intuition behind one of the definitions of a *connected topological space*. Ideally, this is just a little tidbit I'd like to stash in The Back Pocket. But as you can tell already, the length of this post isn't so "little"! Oh well, here we go!

**connected**if it is not

**disconnected,**i.e. if it cannot be written as $X=A\cup B$ where $A$ and $B$ are disjoint, nonempty open subsets of $X$. This seems intuitive enough, right? We can easily visualize a disconnected space. But there's also an equivalent definition:

$X$ is **connected** if every continuous function $f:X\to\{0,1\}$ is constant, where $\{0,1\}$ has the discrete topology*.

*How in the world are these equivalent?*Although the proof isn't too hard, it may be difficult to see the connection (har har)

*intuitively.*But I claim that this second definition of connectedness is just as intuitive as the first! Hopefully by the end of this post, you'll agree.

## What's Really Going on Here?

To say "there is a continuous function from $X$ to $\{0,1\}$" is the mathematician's way of saying "$X$ and $\{0,1\}$ share the same structure." Indeed, what's characteristic about a disconnected space? It's made up of two pieces! (Loosely speaking.) And what's the simplest example of a set which is made up of two pieces? $\{0,1\}$ of course! (Why? Because, well, it's made up of two pieces.) So if we can find a structure preserving map $f$ (a continuous function) from $X$ to $\{0,1\}$ which is *not constant*, then some elements of $X$ are mapped to $0$ while others are mapped to $1$. Hence we can *separate* the elements of $X$ into two subsets by putting a label, either a $0$ or a $1$, on each element. That's what a function does! If you like, imagine that $f$ paints all the elements that map to $0$ red and all the elements that map to $1$ blue. Then we'll have this scenario:

*single*value** in $\{0,1\}$ (or a

*single*color, either red or blue), then it must be that $X$ itself is instrinsically a

*single*unit, that is, $X$ is connected. This observation is really a consequence of the fact that

*continuity preserves connectedness.*You know the theorem:

If $f:X\to Y$ is a continuous map between topological spaces $X$ and $Y$, and if $X$ is connected, then $f(X)$ is connected.

In particular, consider the special case when $Y=\{0,1\}$. If we can find a continuous function $f:X\to\{0,1\}$ such that $f(X)\subset\{0,1\}$ is*not*connected, then by the above we may conclude that $X$ is not connected either. But what does it mean for $f(X)\subset\{0,1\}$ to be "not connected"? It means that $f(X)$ can be written as a union of two

*nonempty*, disjoint, open subsets of $\{0,1\}$. The only* possibility for this is if $f(X)=\{0\}\cup\{1\}$! And this is precisely what it means to say $f:X\to\{0,1\}$ is

*not constant,*which ties back to our illustration above. To put it succintly,

**the connectedness (or disconnectedness) of $X$ is reflected in the connectedness (or disconnectedness) of its image, $f(X)$.**

## Some Final Remarks

*Almost.*" The observation is correct if you replace "continuous functions" by "homeomorphisms." (A homeomorphism is a bijective continuous function whose inverse is also continuous.) Here's the cool thing: the analogy between these two types of maps is no mere coincidence; it's all solidified via the language of

*category theory!*But perhaps we'll save that discussion for another day.

*Footnotes*

* To say "$\{0,1\}$ has the discrete topology" means that the open sets of $\{0,1\}$ are $\varnothing,\{0\},\{1\},$ and $\{0,1\}$.

**This is what it means for $f:X\to\{0,1\}$ to be constant, either all of $X$ gets sent to $0$ or all of $X$ gets sent to $1$.