Compactness of $Y$ implies compactness of $X$

Question

Question is as follows :

Suppose $p:X\rightarrow Y$ is a closed continuous surjection such that $p^{-1}(\{y\})$ is compact for each $y\in Y$. Show that if $Y$ is compact then $X$ is compact.

Hint :

If $U$ is an open set containing $p^{-1}(\{y\})$ there is a neighborhood $W$ of $y$ such that $p^{-1}(W)\subseteq U$.

Let $U$ be an open set in $X$ such that $p^{-1}(\{y\})\subset U$. We then have $y\in p(U)$. Suppose $y\notin p(U)$, because $p$ is surjective there exists $x'\in X\setminus U$ such that $y=p(x')$. So, $x'\in p^{-1}(\{y\})\subset U$ a contradiction. So, $y\in p(U)$.

$U$ is open implies $X\setminus U$ is closed which implies $p(X\setminus U)$ is closed which implies $Y\setminus p(U)$ is closed which means $p(U)$ is open.

So, we have an open set $p(U)$ such that $y\in p(U)$.. If i have to set $W=p(U)$ then i need to have $p^{-1}(p(U))\subseteq U$ which is not true always.. Help me to clear this..

Answer 1

Claim: With $p$ as above we have:

$$ X \space compact \iff Y \space compact$$

Proof:

Since $p: X \to Y$ is a continuous surjection we obtain the "$\implies$" direction.

For the converse pick $U_{\alpha}$ an open cover of $X$. Since $p^{-1}(\{y\})$ is compact and is covered by $\cup_{\alpha} U_{\alpha}$ there is a finite subcover covering $p^{-1}(\{y\})$. We get $p^{-1}(\{y\}) \subset U_{\alpha (y)}:=\cup_{1 \leq i \leq m}U_{\alpha_i}$. Because $p$ is closed we know that $F_{\alpha (y)}:=p(X\setminus U_{\alpha (y)})$ is closed aswell. Therefore $W_{\alpha (y)}:=Y\setminus F_{\alpha (y)}$ is open and $$p^{-1}(W_{\alpha (y)})=p^{-1}(Y \setminus F_{\alpha (y)})=X \setminus p^{-1}(F_{\alpha (y)})=X\setminus p^{-1}(p(X \setminus U_{\alpha (y)})) \subset U_{\alpha (y)}$$

where in the last step we used $A \subset f^{-1}(f(A))$ because there might be more elements mapping into $A$ (I leave it to you to figure this out, maybe draw a picture rather than prove it rigorously).

This actually proves the hint.

Now observe that $$\cup_{y \in Y} W_{\alpha (y)}=\cup_{y \in Y}Y \setminus p(X\setminus U_{\alpha(y)})=Y\setminus \cap_{y \in Y}p(X \setminus U_{\alpha(y)}) \supset Y\setminus p(\cap_{y \in Y}X\setminus U_{\alpha(y)})=Y \setminus p(X\setminus \cup_{y\in Y}U_{\alpha(y)}) \supset Y$$

where in the last step we used the fact that $p^{-1}(\{y\})\subset U_{\alpha(y)}$. (To see the set-theoretic argument of the inclusion draw a picture).

Finally use the compactness of $Y$ to extract a finite subcover $\{W_{\alpha(y)_j}\}_{1\leq j \leq n}$ of $Y$. We then obtain:

$$X=\cup_{1 \le j \le n}p^{-1}(W_{\alpha(y)_j}) \subset \cup_{1 \le j \le n} U_{\alpha(y)_j}$$

Since the $U_{\alpha(y)}$ are just finite unions of sets in the open cover $U_{\alpha}$ we were able to finde a finite subcover of the original (arbitrary) cover. This gives us the compactness of $X$ and concludes the proof.

Answer 2

Suppose that $\{ U_i : i \in I \}$ is an open cover of $X$.

Consider the family $\mathcal V$ of all sets of the form $$Y \setminus p [ X \setminus ( U_{i_1} \cup \cdots \cup U_{i_n} ) ]$$ where $i_1, \ldots , i_n \in I$. It is fairly straightforward to show that this is a family of open subsets of $Y$. Furthermore, given $y \in Y$ as $p^{-1} (y)$ is compact and $\{ U_i : i \in I \}$ covers this set, there are $i_1, \ldots , i_n \in I$ such that $p^{-1}(y) \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$. It follows that $p^{-1}(y) \cap X \setminus ( U_{i_1} \cup \cdots \cup U_{i_n} ) = \emptyset$, and so $y \notin p[ X \setminus ( U_{i_1} \cup \cdots \cup U_{i_n} ) ]$, and so $y \in Y \setminus p[ X \setminus ( U_{i_1} \cup \cdots \cup U_{i_n} ) ]$. That is, $\mathcal V$ covers $Y$.

By compactness of $Y$ there are $V_i , \ldots , V_m \in \mathcal V$ such that $Y = V_1 \cup \cdots \cup V_m$. For each $j \leq m$ there are $i_{j,1} , \ldots , i_{j,n_j} \in I$ such that $$V_j = Y \setminus p [ X \setminus ( U_{i_{j,1}} \cup \cdots \cup U_{i_{j,n_j}} ) ].$$ In particular, only finitely many $i$ appear as a $i_{j,k}$. If $\{ U_{i_{j,k}} : j \leq m, k \leq n_j \}$ does not cover $X$, take $x \in X$ not in the union.

Given $j \leq m$ it follows that $x \notin U_{i_{j,1}} \cup \cdots \cup U_{i_{j,n_j}}$, and so $x \in X \setminus ( U_{i_{j,1}} \cup \cdots \cup U_{i_{j,n_j}} )$, and so $p(x) \in p [ X \setminus ( U_{i_{j,1}} \cup \cdots \cup U_{i_{j,n_j}} ) ]$, and so $p(x) \notin Y \setminus p [ X \setminus ( U_{i_{j,1}} \cup \cdots \cup U_{i_{j,n_j}} ) ] = V_j$. This then contradicts that $V_1, \ldots , V_m$ covers $Y$!

Therefore it must be that $\{ U_{i_{j,k}} : j \leq m, k \leq n_j \}$ covers $X$.

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To use the "family of closed subsets with the finite intersection property" characterisation of compactness, you will likely have to also consider finite subfamilies of this family. To wit:

Suppose that $\mathcal{F}$ is a family of closed subsets of $X$ with the finite intersection property. Note that without loss of generality we may assume that $\mathcal F$ is closed under finite intersections (i.e., if $F_1 , \ldots , F_n \in \mathcal F$, then $F_1 \cap \cdots \cap F_n \in \mathcal F$). (If needed, just replace $\mathcal F$ with the family $\{ F_1 \cap \cdots \cap F_n : F_1 , \ldots , F_n \in \mathcal F \}$. This is clearly also a family of closed subsets of $X$, and as any finite intersection of sets in this family is at heart an intersection of finitely many sets from the original $\mathcal F$, it also has the finite intersection property.)

Consider the family $\{ p[F] : F \in \mathcal F \}$. As $p$ is a closed mapping, this is a family of closed subsets of $Y$. It also follows that this family has the finite intersection property. (Given $F_1 , \ldots , F_n \in \mathcal F$, if you pick $x \in F_1 \cap \cdots \cap F_n$ — remember that $\mathcal F$ has the finite intersection property — then $p(x) \in p[F_1] \cap \cdots \cap p[F_n]$.) By compactness of $Y$ it follows that $\bigcap \{ p[F] : F \in \mathcal F \}$ is nonempty, so pick $y$ in the intersection.

By assumption $p^{-1} \{ y \}$ is a compact subset of $X$, and clearly $\{ p^{-1} \{ y \} \cap F : F \in \mathcal F \}$ is a family of closed subsets of $p^{-1} \{ y \}$. I claim that this family has the finite intersection property.

• Given $F_1 , \ldots , F_n \in \mathcal F$, note that $( p^{-1} \{ y \} \cap F_1 ) \cap \cdots \cap ( p^{-1} \{ y \} \cap F_n ) = p^{-1} \{ y \} \cap ( F_1 \cap \cdots \cap F_n )$. By assumption $F_1 \cap \cdots \cap F_n \in \mathcal F$, and $y \in p[ F_1 \cap \cdots \cap F_n ] \subseteq p[F_1] \cap \cdots \cap p[F_n]$. Thus there is an $x \in F_1 \cap \cdots \cap F_n$ such that $p(x) = y$, and so $p^{-1} \{ y \} \cap ( F_1 \cap \cdots \cap F_n ) \neq \emptyset$.

As $p^{-1} (y)$ is compact, it follows that $\bigcap \{ p^{-1} ( y ) \cap F : F \in \mathcal F \}$ is nonempty, so in particular $\bigcap \{ F : F \in \mathcal F \}$ is nonempty.

Answer 3

Let $\mathcal U$ be an ultrafilter on $X$. Then, $f\mathcal U$ is an ultrafilter on $Y$ and thus $f\mathcal U \to y$. Now, the claim is that for some $x$ in $f^{-1}(y)$, $\mathcal U \to x$. Otherwise, for all $x\in f^{-1}(y)$ we pick an open $V_x$ that is NOT in $\mathcal U$. Then, we use compactness to get a finite subcover $f^{-1}(y) \subset V_1 \cup \cdots \cup V_k$. But then, $X\setminus V_i \in \mathcal U$, and so $C = f(\bigcap X\setminus V_i) \in f\mathcal U$, is closed (since f is closed), and does not contain $y$, but that means $Y\setminus C$ is not in $f\mathcal U$, is open, and contains $y$. This contradiction shows that $\mathcal U \to x$ for some $x\in f^{-1}(y)$. Thus, $X$ is compact. This is actually just a repackaging of the above answers.

Answer 4

Let $\mathcal{C}$ be a collection of closed sets in $X$ having finite intersection property.

Then $\{f(C)\}_{C\in \mathcal{C}}$ is a collection of closed sets in $Y$ as $f$ is closed.

See that $f\left(\bigcap _{i=1}^n\right)\subset \bigcap_{i=1}^n f(C_i)$. So, $\{f(C)\}_{C\in \mathcal{C}}$ has finite intersection property.

Because $Y$ is compact, this collection $\{f(C)\}_{C\in \mathcal{C}}$ has non empty intersection.

Let $y\in \bigcap f(C)$. The set $f^{-1}(\{y\})$ is compact. The collection $\{f^{-1}(\{y\})\cap C : C\in \mathcal{C}\}$ is a collection of closed sets in $f^{-1}(\{y\})$. As $f^{-1}(\{y\})$ is compact, this collection has non empty intersection if this has finite intersection property.

Enough to prove that $f^{-1}(\{y\})\bigcap (\bigcap_{i=1}^n C_i)\neq \emptyset$. Otherwise, we have $f^{-1}(\{y\})\bigcap (\bigcap_{i=1}^n C_i)= \emptyset$ i.e., $f^{-1}(\{y\})\subset (\bigcap_{i=1}^n C_i)^c=\bigcup_{i=1}^n C_i^c$.

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So, $f^{-1}(y)\subset C_i^c$ for some $C_i\in \mathcal{C}$. So,$$y\in f(C_i^c)=f(X\setminus C_i)=f(X)\setminus f(C_i)=Y\setminus f(C_i)=f(C_i)^c$$ But then, we have assumed that $y\in f(C)$ for all $C\in\mathcal{C}$ and above conclusion says $y\notin f(C_i)$ for some $C_i$ a contradiction. So, $f^{-1}(\{y\})\bigcap (\bigcap_{i=1}^n C_i)\neq \emptyset$ for any finite subcollection $\{C_i\}$ in $\mathcal{C}$.

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The justification between above two lines is not correct as $f^{-1}(\{y\})\subset \bigcup_{i=1}^n C_i^c$ need not imply $f^{-1}(y)\subset C_i^c$ for some $i$.

So, we change the collection... We can assume that $\mathcal{C}$ is such that $\mathcal{C}$ is closed under finite union.. This is same as considering the collection $\left\{f\left(\bigcap_{i=1}^n C_i\right)\right\}$.. So, from $f^{-1}(\{y\})\bigcap (\bigcap_{i=1}^n C_i)= \emptyset$ we see that $f^{-1}(\{y\})\bigcap C=\emptyset$ for some $C=\bigcap_{i=1}^n C_i$ in $\mathcal{C}$.... $f^{-1}(y)\subset C^c$ for some $C\in \mathcal{C}$. So,$$y\in f(C^c)=f(X\setminus C)=f(X)\setminus f(C)=Y\setminus f(C)=f(C)^c$$ But then, we have assumed that $y\in f(C)$ for all $C\in\mathcal{C}$ and above conclusion says $y\notin f(C)$ for some $C$ a contradiction. So, $f^{-1}(\{y\})\bigcap (\bigcap_{i=1}^n C_i)\neq \emptyset$ for any finite subcollection $\{C_i\}$ in $\mathcal{C}$.

So, $\{f^{-1}(\{y\})\cap C : C\in \mathcal{C}\}$ collection of closed sets in compact set $f^{-1}(\{y\})$ has finite intersection proeprty. So, $\bigcap (f^{-1}(\{y\})\bigcap C)\neq \emptyset$. So, $\bigcap C\neq \emptyset$ and so $X$ is compact.