The Image of a Connected Space

One theorem turns out to be very important when we’re dealing with connected spaces, or even just with a connected component of a space. If $f$ is a continuous map from a connected space $X$ to any topological space $Y$ , then the image $f(X)\subseteq Y$ is connected. Similarly, if $X$ is path-connected then its image is path-connected.

The path-connected version is actually more straightforward. Let’s say that we pick points $y_0$ and $y_1$ in $f(X)$ . Then there must exist $x_0$ and $x_1$ with $f(x_0)=y_0$ and $f(x_1)=y_1$ . By path-connectedness there is a function $g:\left[0,1\right]\rightarrow X$ with $g(0)=x_0$ and $g(1)=x_1$ , and so $f(g(0))=y_0$ and $f(g(1))=y_1$ . Thus the composite function $g\circ f:\left[0,1\right]\rightarrow Y$ is a path from $y_0$ to $y_1$ .

Now for the connected version. Let’s say that $f(X)$ is disconnected. Then we can write it as the disjoint union of two nonempty closed sets $B_1$ and $B_2$ by putting some connected components in the one and some in the other. Taking complements we see that both of these sets are also open. Then we can consider their preimages $f^{-1}(B_1)$ and $f^{-1}(B_2)$ , whose union is $X$ since every point in $X$ lands in either $B_1$ or $B_2$ .

By the continuity of $f$ , each of these preimages is open. Seeing as each is the complement of the other, they must also both be closed. And neither one can be empty because some points in $X$ land in each of $B_1$ and $B_2$ . Thus we have a nontrivial clopen set in $X$ , contradicting the assumption that it’s connected. Thus the image $f(X)$ must have been connected, as was to be shown.

From this theorem we see that the image of any connected component under a continuous map $f$ must land entirely within a connected component of the range of $f$ . For example, any map from a connected space to a totally disconnected space (one where each point is a connected component) must be constant.

When we specialize to real-valued functions, this theorem gets simple. Notice that a connected subset of $\mathbb{R}$ is just an interval. It may contain one or both endpoints, and it may stretch off to infinity in one or both directions, but that’s about all the variation we’ve got. So if $X$ is a connected space then the image $f(X)$ of a continuous function $f:X\rightarrow\mathbb{R}$ is an interval.

An immediate corollary to this fact is the intermediate value theorem. Given a connected space $X$ , a continuous real-valued function $f$ , and points $x_1,x_2\in X$ with $f(x_1)=a_1$ and $f(x_2)=a_2$ (without loss of generality, $a_1<a_2$ ), then for any $b\in\left(a_1,a_2\right)$ there is a $y\in X$ so that $f(y)=b$ . That is, a continuous function takes all the values between any two values it takes. In particular, if $X$ is itself an interval in $\mathbb{R}$ we get back the old intermediate value theorem from calculus.

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January 3, 2008 - Posted by John Armstrong | Analysis, Calculus, Point-Set Topology, Topology

7 Comments »

Apologies if you covered this in an earlier post, but the (nice) proof you give of the intermediate value theorem relies on knowing that intervals are connected — which is one of those screamingly obvious things that still needs proof. Some variant on the Archimedean axiom, if I recall correctly?

Comment by toomuchcoffeeman | January 4, 2008 | Reply
Above I said “Notice that a connected subset of $\mathbb{R}$ is just an interval”, which is implicitly an invitation to verify it for yourself if you feel it needs verification.

Roughly the proof for a finite closed interval goes by breaking the interval $\left[a,b\right]$ into two open sets $U$ and $V$ with no overlap between them. Then you consider the set of $x$ so that $\left[a,x\right]\subseteq U$ . This set is bounded above by $b$ and so must have a least upper bound $c$ . Then you show that $c$ cannot be in either $U$ or $V$ , and so the interval is connected.

Technically there’s also the matter of showing that a connected subset of $\mathbb{R}$ is an interval, which is the converse of the above.

Comment by John Armstrong | January 4, 2008 | Reply
[...] Your Table with the IVT Here’s a nice, light application of the Intermediate Value Theorem we came up with yesterday. Let’s say you’ve got a square table with four legs at the [...]

Pingback by Stabilizing Your Table with the IVT « The Unapologetic Mathematician | January 4, 2008 | Reply
Above I said “Notice that a connected subset of \mathbb{R} is just an interval”, which is implicitly an invitation to verify it for yourself if you feel it needs verification.

Yes, I wasn’t questioning either the validity or elegance of the IVT proof you gave. I’m a little puzzled as to whether you are using `you’ in the sense of the French `on’ or the german `man’, or not.

(Declaration of interest: as an analyst, I tend towards the view that all the tedious things one learns in Analysis I or Analysis II are obviously true, and easy to prove *once you have the right definitions* — but getting the setup right is non-trivial.)

I look forward to seeing the Ham Sandwich Theorem as a follow up post…

Comment by toomuchcoffeeman | January 4, 2008 | Reply
I mean “you” as in “a generic member of the audience”, yes.

Comment by John Armstrong | January 7, 2008 | Reply
[...] won’t even get a chance to post tomorrow, so I’ll take a cute from a commenter on my Intermediate Value Theorem post and mention the “ham sandwich” [...]

Pingback by The Ham Sandwich Theorem « The Unapologetic Mathematician | January 7, 2008 | Reply
[...] some between the maximum and minimum of on . But then since is connected, we know that there is some so that , as we asserted. Possibly related posts: (automatically generated)The [...]

Pingback by The Geometric Interpretation of the Jacobian Determinant « The Unapologetic Mathematician | January 8, 2010 | Reply

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