Open Continuous Image of Paracompact Space is not always Countably Metacompact

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Theorem

Let $T_A = \left({X_A, \tau_A}\right)$ be a topological space which is paracompact.

Let $T_B = \left({X_B, \tau_B}\right)$ be another topological space.

Let $\phi: T_A \to T_B$ be a mapping which is both continuous and open.


Then it is not necessarily even the case that $T_B$ is countably metacompact, let alone paracompact.


Proof

Let $T_X = \left({X, \tau}\right)$ be a countable discrete space.

Let $T_Y = \left({\left\{{0, 1}\right\}, \tau_0}\right)$ be a Sierpiński space.

Let $T_A = T_X \times T_Y$ be the product space of $T_X$ and $T_Y$.

Then $T_A$ is paracompact from Product of Countable Discrete Space with Sierpiński Space is Paracompact.


Let $T_B = \left({S, \tau_p}\right)$ be a countable particular point space.

Then $T_B$ is not countably metacompact from Infinite Particular Point Space is not Countably Metacompact


Let $\phi: T_A \to T_B$ be a mapping defined as:

$\forall \left({x, y}\right) \in T_A: \phi \left({\left({x, y}\right)}\right) = \begin{cases} p & : y = 0 \\ x & : y = 1 \end{cases}$


Then $\phi$ is open and continuous as follows:



So we have constructed $\phi$ which is both open and continuous from a paracompact space to a space which is not countably metacompact.

$\blacksquare$


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