How to show this function doesn't exist?

Question

Let B be a Baire space, M a metric space and D a subset of B which is meager and dense in B. Show there's no f:B --> M that is continuous precisely at the elements of D.

Based on this fact, conclude there's no f:R^m --> R^n continuous precisely at the elements of.Q^m.

thank you.

Answer 1

I am adapting this proof from Munkres' Topology: A First Course (linked below), pp. 295-296, of a similar proposition restricted to a function f: R→R. It seems to me that the proof carries over; I may be overlooking a detail, however.

The crucial factor is that D is not a G-δ set in B.

Let U_n = union {U: U open in B, diam(f(U)) < 1/n}, and let C = ∩{U_n: n in N}. Then the claim is that C is precisely the set of points of continuity of f. (I leave a proof of this claim to the reader.)

But note that C is a G-δ set in B, whereas D is not; hence D≠C, i.e. D cannot be the entire set of points of continuity of any such function f.
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The conclusion is trivial from the theorem.
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Answer 2

I'd just like to add a few points to Jeredwm's answer.

With a slightly modification of his proof, we can show that, if T is any toplogical space and M is a metric space, then the set of points of T at which any f:T--> M is continuous is a G-delta. Actually, all you have to do is replace the Euclidean norm || || with the distance function d:M x M --> [0, oo) that makes M into a metric space.

In our case, T = B is a Baire space, which implies, in virtue of the definition of a Baire space, that, except for the empty set, no open subset of B is meager. In particular, the whole space B is not meager.

Why isn't the set D a G-delta?

First, we see that B = D U D', with D meager. If both D and D' were meager, so would B (countable - so finite- unions of meager sets are meager), contrarily to the assumption that B is a Baire space. Hence, D' is not meager.

If D were a G-delta, it's complement D' would be F-sigma, so that D' would be given by the union of a countable collection {C_n} of closed sets. Since D is dense in B, D' has an empty interior, which automatically implies that each C_n also has an empty interior. So, each C_n, being closed, is nowhere dense, and D, being given by their union, is meager. But this contradicts the previous conclusion that D' is not meager, which shows the assumption that D is a G-delta is unsustainable.

So, no f:B --> M can have D as the set of their elements of continuity.

Now, specializing this result to R^m and Q^m:

We know that R^m, being a complete metric space according to the Euclidean norm, is a Baire space (complete metric spaces and compact Hausdorff spaces are Baire spaces - Also Spake Baire in his theorem!). Every metric space is a Haursdorff - so a T1- topological space, so that singletons (sets like {a}) are closed. In addition, R^m has no isolated points, which implies singletons have an empty interior, being, therefore, nowhere dense. This, in turn, implies 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 conclude no function from R^m to any metric space can have Q^m as the set of its points of continuity.

Actually, this same conclusion applies to any countable dense set of R^m, like, for example, A^m, where A is the set of the algebraics.

Answer 3

|f(z)| > |z| so in case you chop up it up into 2 equations it turns into -f(z) > -z and -f(z) > -z comparable ingredient now use a attempt element we could say 3 for f of z -f(3) > -3 divide the two aspects by making use of -a million in case you please f(3) > 3 f of z is on no account going to be extra effective than z for this reason it is impossible.