General topology I Bourbaki exercise 15 sec. 3 ch. 1

Question

The exercise reads as follows (only the first part): Consider the following subspaces or $\mathbb R$: $X_1=(-1,1)$, $X_2=(-2,1)$ and $X_3=(1,2)$, and let $X$ be the sum of the $X_i$. Let $A_2$ and $A_3$ be the set of irrational numbers contained in $X_2$ and $X_3$ respectively. Let $X/\mathcal R$ be the space obtained by pasting together the $X_i$ along $A_2$ and $A_3$ by means of the homeomorphism $h(x)=x+3$. Show that the subspace $\mu(X_1)$ of $X/\mathcal R$ is not homeomorphic to $X_1$ $\mu:X\rightarrow X/\mathcal R$ is the quotient map).

To make the exercise 'easier' (or lighter in notation) I think I can translate the spaces since all of them are homemorphic: $X_1=(0,2)$ (to avoid negative numbers), $X_2=(2,5)$ and $X_3=(5,6)$, while the homemorphism $h$ remains unchanged. Then the sum $X$ is just the union of the $X_i$.

Now, the topology on $X/\mathcal R$ is the final topology with respect to the family $\{(X_i,\iota_i\circ \mu)\}_{i=1}^3$. I think $\mu(X_2)$ and $\mu(X_3)$ are not open (neither closed) in $X/\mathcal R$. For example, $\mu^{-1}(\mu(X_2)=X_2\cup Irr(X_3)$; because of the set of irration numbers contained in $X_3$ is not open $\mu(X_2)$ is not open. However, $\mu(X_1)$ should be open in $X\mathcal R$.

Now, the exercise. According to no.5 of section 2, the topology of $X_1$ is strictly finer than the topology induced on $\mu(X_1)$ by $X/\mathcal R$. Then, I should be able to find an open set in $X_1$ which is not open in $\mu(X_1)$. But I can't. For example, if $(a,b)$ is an open interval in $X_1$, then $\mu((a,b))$ is open in $X/\mathcal R$. As a consequence, it is open in $\mu(X_1)$, since it is the intersection of $\mu(X_1)$ with $\mu((a,b))$. I can't imagine any subset $U\subset\mu(X_1)$ that cannot be write as the intersection of $\mu(X_1)$ and an open subset of $X/\mathcal R$.

EDIT Regarding Paul Frost's comment, I think that I have to do the following aclaration: In the general description of the book (sec 2 no 5) to each $X_i$ there corresponds a subset $A_i $. But in this exercise there are only two sets indicated: $A_2$ and $A_3$. Then I have set $A_1=\emptyset$. Then, for me the equivalence class of each element of $x\in X_1$ contains only $x$ itself.

EDIT2:

Once @PaulFrost indicated me how the equivalence relation $\mathcal R$ I have been able to elaborate my own answer. I would like you to check it.

The motivation of this exercise is to show that the topology (call it $\tau$) defined on $X_1'=\mu(X_1)$, $\mu$ is the quotient map, by means of $X_1$ and the restriction $\mu|_{X_1}$ which is a bijection, is strictly finer than the topology on $X_1'$ induced by the quotient $X/\mathcal R$ (call it $\tau'$). If I show that I will solved my problem.

Now, I want to prove that there are open sets in $X_1'$ in $\tau$ that are not open in $\tau'$. In particular, I focus on the open intervals of the form $(a-\epsilon,a+\epsilon)$, with $a<1$ and $\epsilon$ ''very small'' ($0<a-\epsilon<a+\epsilon<1$). Let $I$ be an interval of this type. $I$ is not open in the quotient space because $\mu^{-1}(\mu(I))= I\cup Rat((a+2-\epsilon,a+2+\epsilon))$.

However, it may exist an open set $U\subset X/\mathcal R$ such that $X_1'\cap U=\mu(I)$. Obviously $\mu(I)\subset U$ and hence, $U$ must contain $\mu((a+2-\epsilon,a+2+\epsilon))$ to be open; otherwise $\mu^{-1}(U)\cap X_2$ would not be open. But by the same reason, $U$ must contain also $\mu((a+5-\epsilon,a+5+\epsilon)$. And this forces $U$ to contain $\mu((a+1-\epsilon,a+1+\epsilon))$. Summarizing: an open set in $X/\mathcal R$ which contains $\mu(a-\epsilon,a+\epsilon)$ must contain the set

$$ \mu\Big((a-\epsilon,a+\epsilon)\cup(a+1-\epsilon,a+1+\epsilon)\cup(a+2-\epsilon,a+2+\epsilon)\cup(a+5-\epsilon,a+5+\epsilon) \Big) .$$

But then

$$ X_1'\cap U = \mu\Big((a-\epsilon,a+\epsilon)\cup(a+1-\epsilon,a+1+\epsilon)\Big) \neq \mu\Big((a-\epsilon,a+\epsilon)\Big). $$

This proves that the searched $U$ does not exists and thus $\mu((a-\epsilon,a+\epsilon))$ is not open in $X_1'$ in $\tau'$.

What do you think? Is my proof valid?

Answer 1

You proof is correct. I nevertheless suggest some "streamlining".

First, you should work with the original $X_i$. Next, consider an open $U_1 \subset X_1$ which contains an interval $(a,b)$ with $0 \le a < b \le 1$ and check what it must look like in order that $p(U_1)$ is open in $X'_1$.

So assume that $p(U_1)$ is open in $X'_1$. Then $p(U_1) = U \cap X'_1$ with an open $U \subset Q$. Therefore $p^{-1}(U)$ is open in $X$. But $p(C_3 \cap (a+1,b+1)) = p(C_1 \cap (a,b)) \subset p(U_1) \subset U$, thus $C_3 \cap (a+1,b+1) \subset p^{-1}(U)$. This implies $(a+1,b+1) \subset p^{-1}(U)$. Hence $p(A_2 \cap (a-2,b-2)) = p(A_3\cap (a+1,b+1)) \subset U$ and therefore $A_2 \cap (a-2,b-2)) \subset p^{-1}(U)$. This implies $(a-2,b-2) \subset p^{-1}(U)$. Hence $p(B_1 \cap (a-1,b-1)) = p(B_2 \cap (a-2,b-2)) \subset U$, thus $B_1 \cap (a-1,b-1) \subset p^{-1}(U)$. This implies $(a-1,b-1) \subset p^{-1}(U)$. But $(a-1,b-1) \subset X_1$ and $U_1 = p^{-1}(U) \cap X_1$ which implies $(a-1,b-1) \subset U_1$.

This means that $p_1 : X_1 \to X'_1$ is no homeomorphism, since $p((a,b))$ is never open when $(a,b) \subset (0,1)$. However, as I explained, this does not suffice to show that $X_1$ are $X'_1$ are not homeomorphic. To make it more transparent what the problem is, consider the following example.

Let $Y = \bigcup_{n=0}^\infty [2n,2n+1) \subset \mathbb{R}$. Let $\tau$ be the subspace topology inherited from $\mathbb{R}$ and $\tau' = \tau \cup \sigma$, where $\sigma$ is the subspace topology of $[\frac{1}{2},1) \subset \mathbb{R}$. It is easy to see that $\tau'$ is a topology on $Y$ which is strictly finer than $\tau$. However, $(Y,\tau)$ and $(Y,\tau')$ are homeomorphic. In fact, define

$$h : Y \to Y, h(t) = \begin{cases} \frac{t}{2} & t \in [0,1) \\ \frac{t-2}{2} + \frac{1}{2} & t \in [2,3) \\ t-2 & t \in [2n,2n+1), n > 1 \end{cases}$$

$h$ is a bijection and it is readily verifed that $h : (Y,\tau) \to (Y,\tau')$ is a homeomorphism. In fact, $\tau'$ was defined so that $\tau' = h(\tau)$.

The proof that $X_1 \not\approx X'_1$ can be done as in my first answer. However, your approach can be used to show that $X'_1$ is not Hausdorff which provides an alternative proof.

Assume $X'_1$ is Hausdorff. Then the points $p(\pm\frac{1}{2})$ can be separated by disjoint open $U'_\pm \subset X'_1$. Since $p_1 : X_1 \to X'_1$ is continuous, the sets $U_\pm = p_1^{-1}(U'_\pm)$ are disjoint open neighborhoods of $\pm\frac{1}{2}$ in $X_1$. Choose a subinterval $(a,b) \subset (0,1)$ such that $\frac{1}{2} \in (a,b) \subset U_+$. Since $p(U_+) = U'_+$ is open in $X'_1$, we know that $-\frac{1}{2} \in (a-1,b-1) \subset U_+$. But this means $U_+ \cap U_- \ne \emptyset$ which is a contradiction.

Not that this argument can be generalized to show that $p(x_1), p(x_2)$ have disjoint open neighborhods in $X'_1$ if and only if $\lvert x_1 - x_2 \rvert \ne 1$.

Answer 2

The three "pastings" give us three relations $\sim_\nu$ on the sum $X = X_1 \cup X_2 \cup X_3$. Explictly we have

$a_2 \sim_{(i)}(a_2) h_{(i)}(a_2) = a_2 + 3$ for $a_2\in A_2$

$b_1 \sim_{(ii)} h_{(ii)}(b_1) = b_1 - 1$ for $b_1 \in B_1$

$c_1 \sim_{(iii)} h_{(iii)}(c_1) = c_1 + 1$ for $c_1 \in C_1$

Let $\mathcal{R}$ denote the equivalence relation on $X$ generated by these $\sim_\nu$ and let $p : X \to X/\mathcal{R} = Q$ denote the quotient map. It is obviuos that $p$ does not identify distinct points within any $X_i$ because the $h_\nu$ are injective. Hence $p$ establishes a bijection between $X_i$ and $X'_i = p(X_i) \subset Q$, but clearly $X'_1 \cap X'_2 = p(B_1) = p(B_2) = B \ne \emptyset$, $X'_1 \cap X'_3 = p(C_1) = p(C_3) = C \ne \emptyset$ and $X'_2 \cap X'_3 = p(A_2) = p(A_3) = A \ne \emptyset$. It is easy to see that $X'_1 \cap A = \emptyset$. Moreover $$Q = p(X_1) \cup p(X_2) \cup p(X_3) = p(X_1) \cup p(X_2 \setminus B_2) \cup p(X_3\setminus C_3) = $$ $$p(X_1) \cup p(A_2) \cup p(A_3) = X'_1 \cup A .$$

Let us see how the topology on $Q = X'_1 \cup A$ looks like. Write $U \subset Q$ as $U = U'_1 \cup U'$ with $U'_1 \subset X_1, U' \subset A$. There exist unique $U_1 \subset X_1, U_2 \subset A_2$ such that $p(U_1) = U'_1, p(U_2) = U'$. We have $$p^{-1}(U) = p^{-1}(p(U_1)) \cup p^{-1}(p(U_2)) = U_1 \cup h_{(ii)}(U_1 \cap B_1)\cup h_{(iii)}(U_1 \cap C_1) \cup U_2 \cup h_{(iii)}(U_2) .$$ Now $U$ is open in $Q$ if and only if $p^{-1}(U)$ is open in $X$. The latter means that

1) $U_1$ is open in $X_1$

2) $V_ 2 = h_{(ii)}(U_1 \cap B_1) \cup U_2$ is open in $X_2$

3) $V_3 = h_{(iii)}(U_1 \cap C_1) \cup h_{(i)}(U_2)$ is open in $X_3$.

Assume that $U_1 \subset (0,1)$. Then $U_1 \cap B_1 = \emptyset$ and we get $V_2 = U_2 \subset A_2$ which possible only when $U_2 = \emptyset$. Thus $V_3 = h_{(iii)}(U_1 \cap C_1)$ which is possible only when $U_1 = \emptyset$. This shows that $p_1 : X_1 \stackrel{p}{\rightarrow} X'_1$ is no homeomorphism.

Now let us assume that there exists a homeomorphism $h : X'_1 \to X_1$. Then $h' = h \circ p_1 : X_1 \to X_1$ is a continuous bijection. This is possible only if $h'$ is strictly increasing or strictly decreasing. W.l.o.g. assume it is strictly increasing. Then $\lim_{t \to -1} h'(t) = -1$ and $\lim_{t \to 1} h'(t) = 1$. This shows that $h'$ has a continuous extension $H : [-1,1] \to [-1,1]$ such $H(\pm1) = \pm1$. Therefore $H$ is a continuous bijection between compact spaces and thus a homeomorphism. We conclude that $h'$ is also one. This implies that $p_1 = h^{-1} \circ h'$ is a homeomorhism which is not true.

Therefore $X_1$ and $X'_1$ are not homeomorphic.