Topology - does this functions exist?

Question

Let B be a Baire space without isolated points that satisfies the T1 axiom. Suppose B contains a countable and dense subset D with complement D'. Can there exist a continuous f:B --> B that sends elements of D into elements of D' and vice versa? Specialize your conclusion for the case of R, Q and I. In this case I know such function doesn't exist, but the proof I know is based on cardinality, not Baire spaces.

Thank you for any help.

Answer 1

No. For the good or for the evil, there’s no function like that.

Proof:

First, let's prove the following lemma:

The set D' is not an F-sigma.
Proof: Since B is a Baire space, non empty open sets are not meager. Therefore, B is not meager.
Since B is T1, one-point sets are closed. And since B has no isolated points, then, for every x in B, any neighborhood of {x} contains an element of B distinct from x, which shows no neighborhood of x is contained in {x}. So, {x} is not open and has an empty interior.
If A is any countable subset of B, then A = {a_1} U {a_2}...U {a_n}.. that is, A is given by a countable union of one-point sets. Since these sets are closed and have empty interior (are nowhere dense), A is meager. Every countable set of B is meager. Therefore, D is meager.
Since B = D U D', with B not meager and D meager, it follows D' is not meager (otherwise, B would be meager, contradicting the assumption that it's a Baire space. We know countable unions of meager sets are meager). And since is D is dense in B, D' has an empty interior (complements of dense sets have empty interiors).
Now, suppose D' is F-sigma. Then, there exists a countable collection {C_n} of closed sets such that
D' = Union (n=1, oo) C_n . Since D' has an empty interior, the same is trivially true for each C_n. So, D' is given by the union of a countable collection of closed sets with an empty interior, that is, nowhere dense sets. This means that, contrarily to our previous conclusion, D' is meager. From this contradiction, it follows D' is not F-sigma and the lemma is proved.

Now, let's go back to the original problem, Suppose f: B --> B is continuous and sends elements of D into elements of D' and vice-versa. Then, f^(-1) (D) , the inverse image of D under f, is the set D'. Since D is countable, D = {d_1} U {d_2}...U {d_n}.....where d_1, d_2...d_n.. form an enumeration of D. As we've seen, each {d_n} is closed. And since f is continuous, the inverse image of each {d_n} is closed in B. By the properties of inverse images, it follows that D' = f^(-1)({d_1}) U f^(-1)({d_2})....U f^(-1)({d_n})...., that is, D' is given by a countable union of closed sets. So, D' is F-sigma, contradicting the lemma we've just proved. From this contradiction, it follows there's no continuous function that sends elements of D into elements of D' and elements of D' into elements of D.

The case R, Q, I is now an immediate consequence of what we have proved. R is a Baire space, since it's a complete metric space (Baire Theorem: Complete metric spaces and compact Hausdorff topological spaces are Baire spaces). R has no isolated points and being a metric space, it's T1, singletons are closed. Q is countable and dense in R. So, we have a particular case of the general theorem we've proved. It follows there's no continuous function from R to R that sends rationals into irrationals and irrationals into rationals. There is indeed, another classical proof of this based on cardinality. Another simple and interesting proof is to set g(x) = f(x) - x and prove that (1) g only takes on irrational values (2) g must be constant. Then, f(x) = x + k for some irrational k. Let x = k, so that f(x) = 2k, irrational -> contradiction.

The theorem you stated also shows that no continuous function from R to R can send algebraics into transcendents and transcendents into algebraics. It's kinda easy to see why, isn't it?

Steiner
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Answer 2

i could in basic terms choose for to characteristic some factors to Jeredwm's answer. With a marginally exchange of his information, we can instruct that, if T is any toplogical area and M is a metric area, then the set of things of T at which any f:T--> M is non-end is a G-delta. incredibly, all you will possibly desire to do is replace the Euclidean norm || || with the area function d:M x M --> [0, oo) that makes M right into a metric area. In our case, T = B is a Baire area, meaning, in distinctive characteristic of the definition of a Baire area, that, aside from the empty set, no open subset of B is meager. in specific, the full area B isn't meager. Why isn't the set D a G-delta? First, we see that B = D U D', with D meager. If the two D and D' have been meager, so could B (countable - so finite- unions of meager instruments are meager), contrarily to the theory that B is a Baire area. subsequently, D' isn't meager. If D have been a G-delta, it is supplement D' could be F-sigma, so as that D' could take transport of by utilising the union of a countable sequence {C_n} of closed instruments. when you consider that D is dense in B, D' has an empty indoors, which promptly means that each and every C_n additionally has an empty indoors. So, each and every C_n, being closed, is nowhere dense, and D, being given by utilising their union, is meager. yet this contradicts the previous end that D' isn't meager, which shows the theory that D is a G-delta is unsustainable. So, no f:B --> M could have D using fact the set of their components of continuity. Now, specializing this result to R^m and Q^m: all of us be conscious of that R^m, being an entire metric area in accordance to the Euclidean norm, is a Baire area (finished metric areas and compact Hausdorff areas are Baire areas - additionally Spake Baire in his theorem!). each and every metric area is a Haursdorff - so a T1- topological area, so as that singletons (instruments like {a}) are closed. besides, R^m has no remoted factors, meaning singletons have an empty indoors, being, as a result, nowhere dense. This, in turn, implies each and every countable set of R^m is meager. besides being countable, so meager, Q^m is dense in R^m. So, the previous result applies and we end no function from R^m to any metric area could have Q^m using fact the set of its factors of continuity. incredibly, this comparable end applies to any countable dense set of R^m, like, as an occasion, A^m, the place A is the set of the algebraics.