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Let $X, Y$ be two metric spaces and $f$ be a continuous bijection (i.e. one-to-one map) from $X$ to $Y$. Let $E$ be a $G_{\delta}$ subset of $X$. I want to know weather the image $f(E)$ is also a $G_{\delta}$ subset of $Y$. Thanks!

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3 Answers 3

up vote 12 down vote accepted

Yes, if $X$ is $\sigma$-compact: $X\smallsetminus E$ is $F_\sigma$, hence a countable union of compact sets. Since a continuous image of a compact set is compact (and therefore closed, in a Hausdorff space), $f(X\smallsetminus E)$ is also $F_\sigma$, hence $f(E)$ is $G_\delta$.

EDIT: While ljjpfx seems to be happy with this answer as his or her spaces are compact, Nate Eldredge asked in a comment whether there is a counterexample with both spaces Polish, which I think is an interesting question. I believe the following example should work. Consider the function $$f\colon\omega^\omega\to[0,1),\quad f(a_0,a_1,a_2,\dots)=0{.}\underbrace{1\cdots1}_{a_0}0\underbrace{1\cdots1}_{a_1}0\underbrace{1\cdots1}_{a_2}0\cdots,$$ where the right-hand side is meant to be the binary representation. (I will write $0{.}1^{a_0}01^{a_1}0\cdots$ to simplify the notation.) It is easy to see that $f$ is a continuous bijection, while $f^{-1}$ is not continuous. Let $U$ be the open set $$\{a=(a_n)_{n\in\omega}\in\omega^\omega:\exists n\\,a_n\text{ is even}\},$$ I claim that $f(U)$ is not $G_\delta$. Assume for contradiction that $f(U)=\bigcap_nG_n$ for some open $G_n\subseteq[0,1)$. Note that $f(U)$ is the union of intervals of the form $[a2^{-k},(a+1)2^{-k})$, where $a,k\in\omega$, and $a$ has the binary representation $1^{a_0}01^{a_1}0\cdots01^{a_r}0$, where $a_i$ are odd for $i< r$, $a_r$ is even, and $k=\sum_{i\le r}(a_i+1)$. Since $G_n\supseteq f(U)$ is open, for every $n,k$ there exists $l_{n,k}$ such that $$G_n\supseteq(a2^{-k}-2^{-(k+l_{n,k})},(a+1)2^{-k})$$ for every $a=1^{a_0}01^{a_1}0\cdots01^{a_r}0$ as above. Note that if $a_r>0$, the end-point of this interval has the form $0{.}1^{a_0}01^{a_1}0\cdots01^{a_{r-1}}01^{a_r-1}01^{l_{n,k}+1}$. Choose $a=(a_n)\in\omega^\omega$ so that every $a_n$ is odd, and $a_{n+1}>l_{n,k_n}$, where $k_n=1+\sum_{i\le n}(a_i+1)$. Then $a\notin U$, but the description above implies that $f(a)\in\bigcap_nG_n$, a contradiction.

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A higher-level argument why $f(U)$ is not $G_\delta$ can be made as follows. Let $D$ be the set of the lower end-points $a2^{-k}$ of the intervals comprising $f(U)$, as described above. Then $P=\overline D$ is a Polish space. Since $D=P\cap f(U)$, $D$ is $G_\delta$ as a subset of $P$, and therefore comeager. However, being a countable set with no isolated points, $D$ is also meager, contradicting the Baire category theorem for $P$. –  Emil Jeřábek Nov 16 '12 at 12:40

No, you have stated nowhere near enough conditions to achieve this.

For example, let $X$ and $Y$ both be your favourite metric space ($R$, (for example) but give $X$ the discrete metric (with $d(x,y)=1$ if $x\neq y$) and let $f$ be the identity (on points).

Then all of the subsets of $X$ are open and so also $G_\delta$. However, they need not have this property in $Y$.

Edit: I have removed, reconsidered and consolidated various other comments that I made about this problem.

My initial reaction was that ljjpfx had omitted to say that $f$ should be a homeomorphism (have a continuous inverse) and not just be continuous and bijective. There are well known conditions under which a continuous bijection $f:X\to Y$ is a homeomorphism, for example if $X$ is compact and $Y$ is Hausdorff.

I thought that all of this would render the original question rather trivial and likely to be closed, whereas I wanted to have the opportunity to develop it into something more interesting.

What I hadn't appreciated is that most subspaces that are defined in any reasonable fashion are $G_\delta$. It seems that Nate didn't notice this either and nor am I convinced by Emil's counterexample.

What is an example of a non-$G_\delta$ subspace of $\Bbb R$?

The direction in which I had wanted to develop this question was by introducing overt subspaces. I asked ljjpfx what situation had given rise to the question because I was wondering whether s/he was actually working with overt subspaces rather than $G_\delta$ ones.

Formally, overtness is the lattice dual of compactness, in particular the direct image of an overt subspace is overt just as happens in the compact case.

The reason why you haven't heard of this notion before is that every (sub)space is overt in classical topology. (The points mess things up, rather than excluded middle.)

The idea comes from various approaches to unifying topology with computation. Overt in English means explicit. In constructive topology it means having evidence.

The Bishop school of constructive analysis has used notions of locatedness and total boundedness to fill in constructive gaps in classical theorems.

Locale Theory and Formal Topology both develop the subject in terms of open subspaces instead of points, the former in a categorical setting (in a topos) and the latter in Martin-Löaut;f type theory. Peter Johnstone and André Joyal called a locale open if its terminal projection is an open map. Giovanni Sambin introduced positivity for formal topology.

My own approach to construcive/computable topology is called Abstract Stone Duality. It started from categorical ideas but developed into a language that only expresses computable continuous functions. My paper A Lambda Calculus for Real Analysis is the best place to start with this; it emphasises the dual behaviour of overt and compact subspaces and applies them to the (constructive) intermediate value theorem.

Recently I have been working on a characterisation of overt subspaces in locally compact metric spaces and open maps between them. I will add a link from this page when I am ready to circulate a draft of this.

In particular, overt subspaces are $G_\delta$, which is why I got involved in this question. However, I now realise that all of the subspaces that I construct in ASD are $G_\delta$ too.

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Is there a counterexample with $X,Y$ both Polish? –  Nate Eldredge Nov 13 '12 at 14:13
Many thanks. I forgot to say both X and Y are compact metric spaces. –  ljjpfx Nov 13 '12 at 15:29
In case it’s not obvious, $\mathbb Q$ is an example of a non-$G_\delta$ subspace of $\mathbb R$. –  Emil Jeřábek Nov 15 '12 at 11:07

No; consider any bijection from $\mathbb{N}$ to $\mathbb{Q} \subset \mathbb{R}$.

Edit: I missed where the question said "bijection"; I saw the next sentence said "one-to-one map" and just thought "injection". Any bijection from $\mathbb{N} \to \mathbb{Q}$ is a continuous injection $\mathbb{N} \to \mathbb{R}$ whose image is $\mathbb{Q}$, which is not $G_\delta$. But of course it is not onto. So this doesn't really answer the question, but I will leave it anyway.

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I don´t quite understand how this answers the question. Wouldn´t you need a continuous bijection from $\mathbb{R}$ onto itself sending $\mathbb{N}$ onto $\mathbb{Q}$? But there is no such a thing, unless you change the topology of your $\mathbb{R}$´s as in Paul´s answer. –  Ramiro de la Vega Nov 13 '12 at 12:16
I like this example on the grounds that $\Bbb Q$ is the one set on which it is natural to put two different topologies (the discrete and Euclidean ones). Unfortunately, I think that every subset of $\Bbb Q$ is $G_\delta$ in the Euclidean topology (at least classically). –  Paul Taylor Nov 13 '12 at 13:20
Thank you all of you. –  ljjpfx Nov 13 '12 at 15:30

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