Set module /blog as homepage

The answer is "Yes, but..." -- You will probably need to tell the software RAID system (mdadm) that the drive has been replaced, following steps similar to these (I wasn't able to find a more ubuntu-specific way of doing this, but you may be able to do it through gnome-disk-utility as well).

There exists a topology $\tau$ with two non-isolated points on a countable set $X$ such that the poset $T_2(\tau)$ does not have minimal elements.

To construct such topology $\tau$, take any Hausdorff $(\omega_1,\omega_1)$-gap on $\omega$, which is a pair $\big((A_\alpha)_{\alpha\in\omega_1},(B_\alpha)_{\alpha\in\omega_1}\big)$ of families of infinite subsets of $\omega$ satisfying the following two conditions:

(H1) for any $\alpha<\beta<\omega_1$ we have $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$;

(H2) for any set $C\subset\omega$ one of the sets $\{\alpha\in \omega_1:A_\alpha\subset^* C\}$ or $\{\alpha\in\omega_1:C\subset^* B_\alpha\}$ is at most countable.

Here the notation $A\subset^* B$ means that the complement $A\setminus B$ is finite.

It is well-known that Hausdorff $(\omega_1,\omega_1)$-gaps do exist in ZFC.

Let $\tau$ be the topology on the countable set $X=\omega\cup\{-\infty,+\infty\}$ consisting of sets $U\subset X$ such that

$\bullet$ if $-\infty\in U$, then there exists $\alpha\in\omega_1$ such that $B_\alpha\subset^* U$;

$\bullet$ if $+\infty \in U$, then there exists $\alpha\in\omega_1$ such that $\omega\setminus A_\alpha\subset^* U$.

It is clear that $-\infty$ and $+\infty$ are unique non-isolated points of the topological space $(X,\tau)$.

We claim that no topology $\sigma$ in the poset $T_2(\tau)$ is minimal.

Since $\sigma$ is Hausdorff, the points $-\infty,+\infty$ have disjoint neighborhoods $U_-,U_+\in\sigma$.

By the condition (H2), one of the sets $A=\{\alpha\in\omega_1:A_\alpha\subset^* U_-\}$ or $B=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$ is countable.

If the set $A$ is countable, then we can find a countable ordinal $\alpha$ such that $A_\alpha\not\subset^*U_-$.

Consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:

$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\cup A_\alpha\subset^* W$;

$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\subset^* W$.

It is clear that $\tau\subset\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ as $U_-\in\sigma\setminus\sigma'$.

The topology $\sigma'$ is Hausdorff since $U_-\cup A_\alpha$ and $U_+\setminus A_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.

If the set $B$ is countable, then we can find a countable ordinal $\alpha$ such that $U_-\setminus\{-\infty\}\not\subset^* B_\alpha$ and hence $\omega\setminus B_\alpha\not\subset^* U_+$.

In this case we can consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:

$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\subset^* W$;

$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\cup(\omega\setminus B_\alpha)\subset^* W$.

It is clear that $\tau\subset\sigma'\subsetneq\sigma$.

The topology $\sigma'$ is Hausdorff since $U_-\cap B_\alpha$ and $U_+\cup(\omega\setminus B_\alpha)$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.

In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$ in $T_2(\tau)$ witnessing that the topology $\sigma$ is not minimal.

Let's examine the manufacturing steps for a very simple single-sided PCB. ¹

PCB fabrication is a subtractive process. The PCB starts life as a solid sheet of copper laminated on a sheet of insulator ². Most of the time, traces are formed by etching. But how to etch away the excess copper between the traces, while keeping the copper that forms the traces?

1. Copper is covered with photoresist.

2. Parts of the board that will remained covered by copper (e.g. traces) are exposed to light.

3. The board is washed with a solvent. Unexposed photoresist is washed away. Exposed photoresist remains. So we've got a board [still] completely covered with copper. Traces are drawn on top of the copper with photoresist.

4. The actual etching takes place.

   • The etching solution attacks copper.
   • At the same time, the chemistry of the photoresist is such that the etching solution doesn't attack it. As a result the traces that are made of copper aren't etched away.
5. Exposed photoresist is washed away with another solvent (different than the one used for unexposes photoresist), which doesn't attack copper.

Thus, we get a board with copper traces.

¹ _{This question is about traces, and I'd like to focus on the traces. The sequence of steps would be somewhat different when there are have multiple copper layers and vias have to be plated.}

² _{Typically fiberglass. If the PCB has to be extra cheap, phenolic paper can be used .}

³ _{Sometimes during prototyping traces can be made by milling.}