1

This is something I came up with while trying to solve another problem; it's a basic point-set topology problem that seems true and easy to prove, but I keep getting stuck.

Let $A,B$ be subspaces of a topological space $X$. Suppose that $C$ is a (connected) component of $A$ which is disjoint from $B$. Suppose moreover that $B,C$ are both closed in $X$. Then $C$ is a component of $A \cup B$.­

Intuitively, the fact that $B,C$ are closed and disjoint prevents the addition of $B$ to interact with the maximal connectivity of $C$, even though $B$ might perturb other components of $A$.

Some thoughts: connectivity of $C$ is clear, as its topology as a subspace of $A \cup B$ coincides with its topology as a subspace of $A$. Now, let $C \subseteq E \subseteq A \cup B$ be a component. If $E$ doesn't intersect $B$, then $E$ may be viewed as a component of $A$ and therefore $E = C$. The more general case where $E$ intersects $B$ and other parts of $A$ is giving me trouble. We may decompose $$E = (E \cap C) \sqcup (E \cap B) \sqcup (E \cap (A \setminus (C \cup B))).$$

If, for instance, $C \cup B$ is open in $A \cup B$, then in the decomposition above all the terms are closed, yielding a disconnection of $E$. But we shouldn't need that assumption.

Alex Provost
  • 20,991
  • You may want to read http://math.stackexchange.com/questions/22032/if-c-is-a-component-of-y-and-a-component-of-z-is-it-a-component-of-y-cup and update your intuition. – Niels J. Diepeveen Jul 21 '16 at 22:55
  • @NielsDiepeveen Thank you. There seem to be many nice results in there, but the common hypothesis seems to be that $C$ is a component of two spaces, which isn't the case here. – Alex Provost Jul 21 '16 at 23:00
  • I have not checked it very thoroughly, but I think that if you take $Y=A, Z = C \cup B, X = Y\cup Z$ or something like that, it is essentially the same. – Niels J. Diepeveen Jul 21 '16 at 23:18

1 Answers1

1

It is probably true for compact Hausdorff spaces, but false in general.

Consider the Knaster-Kuratowski fan $X$. Let $B=\{p\}$ be the dispersion point, and $A=X\setminus B$. Let $a\in A$. Then the component $C$ of $a$ in $A$ is equal to $\{a\}$. So $B$ and $C$ are both closed. but the component of $a$ in $X$ is equal to all of $X$.

Something like this could not happen in compact Hausdorff space because the component of $C$ in $X$ would have to intersect $B$ if $A$ and $B$ are disjoint and $A$ is open (this is sometimes called boundary bumping theorem in texts). It may be interesting to try and prove for compact Hausdorff when $A$ and $B$ are not necessarily disjoint.

Forever Mozart
  • 9,294
  • Thanks; what a fascinating space! I think this is a great example of how basic intuition can fail us in a surprising manner. – Alex Provost Jul 21 '16 at 23:27
  • @AlexProvost yes, it always helps to have a lot of pathological spaces in your head – Forever Mozart Jul 21 '16 at 23:39
  • Do you have any idea if the claim remains false if $X$ itself is less pathological? For the application I had in mind, $X = \mathbb{R}^2$, for instance. – Alex Provost Jul 21 '16 at 23:39
  • my intuition is that it is true, because $\mathbb R^2$ is locally compact (the boundary bumping theorem holds in these spaces), but I'll have to think about it – Forever Mozart Jul 21 '16 at 23:40
  • Thank you, I would love to know if this holds for such a class of spaces. – Alex Provost Jul 21 '16 at 23:41
  • @AlexProvost oh, of course. If $C$ is a component of $A$ that misses $B$, then $C$ is a component of $A\setminus B$, which is open in $X$. By the boundary bumping theorem, the component of $C$ in $X$ intersects $B$. This theorem appears in most books on continuum theory, for instance, theorem 12.10 here: https://books.google.com/books?id=TWo2h710HysC&pg=PA101&lpg=PA101&dq=boundary+bumping+theorem+nadler&source=bl&ots=ji6nr83eCt&sig=zdCzIIdws4cKYKekdSojNTYQhZI&hl=en&sa=X&ved=0ahUKEwiQqPyZ34XOAhUD5yYKHUn8AIgQ6AEIMDAE#v=onepage&q=boundary%20bumping%20theorem%20nadler&f=false – Forever Mozart Jul 21 '16 at 23:49
  • continuum means compact + connected. But it works in locally compact + connected too. You can prove this by taking a compactification of $X$ and then use the bbt with the fact that $X$ is open in the compactification. – Forever Mozart Jul 21 '16 at 23:51
  • Let us continue this discussion in chat. – Alex Provost Jul 21 '16 at 23:57