How to show the set is open/closed

[P3X-018]

I have the set

$$S = \left\{ (a,\ldots,f) \in\mathbb{R}^6\,|\, \vert ad-bc\vert = 0 \,\wedge \, \vert af-be\vert > 0 \right\}$$

Now writting it as (it is possible to split like this, right?)

$$S = \left\{ (a,\ldots,f) \in\mathbb{R}^6\,|\, \vert ad-bc\vert = 0 \right\} \cap \left\{ (a,\ldots,f) \in\mathbb{R}^6\,|\, \vert af-be\vert > 0 \right\}$$

Obviously the set with $\vert af-be\vert > 0$ is an open set in $\mathbb{R}^6$ because of the sharp inequality, and the other closed. But is this enough of an argument to conclude wether the sets are open or closed? Or what more is needed to conclude that? It seems kinda overkill to argue with the definition of an open and closed set, that is to argue that if a set is open then for any point in the set there exists 'sphere' with center at that point contained completely in the set.

 

[arildno]

Suppose you are on a point within S.

1. Are you still within S if you make an arbitrarily tiny change of d, the other components being unchanged?

2. What can you conclude from this?

3. Are you finished, or do you need to prove something more?

 

[P3X-018]

If $a\neq 0$, then by making a small change in d you'll fall outside the first part of S. So that set can't be open, since there is points arbitrarly close to the set that isn't contained in it.
But how can this be used to conclude anything if a = 0?

And what about the other part of S, that is open. You argue the same way as above about that it's complement is closed? That is saying as since $\vert af-bc\vert = 0$ is closed it's complement is open? So this is enough?

 

[arildno]

If $a\neq 0$, then by making a small change in d you'll fall outside the first part of S. So that set can't be open, since there is points arbitrarly close to the set that isn't contained in it.
But how can this be used to conclude anything if a = 0?

Irrelevant!
In order for a set to be open, then it must be possible around EVERY point in the set to construct a tiny ball that lies wholly within the set.

Since with a not equal to 0 we find a COUNTER-EXAMPLE to the requirement for a set to be open, we have proven that the set itself is NOT open.

And what about the other part of S, that is open. You argue the same way as above about that it's complement is closed? That is saying as since $\vert af-bc\vert = 0$ is closed it's complement is open? So this is enough?

What you need to prove now, is whether the set is either closed or not closed (as well as being not open).
Look at the complement of S.
If you can prove that this is open, then S is closed; if the complement of S is NOT open, then S is neither open or closed.