Let E⊂ S , both open sets. Is Ssetminus E necessarily open?

Let $E\subset S$, both open sets. Is $S\setminus E$ necessarily open?

Step 1 · Answer

$$ \boxed{\text{No — } S\setminus E \text{ need not be open}} $$

Step 2 · A counterexample

Let $S=\{z:|z|<2\}$ and $E=\{z:|z|<1\}$ — both open, and $E\subset S$. Then:

$$ S\setminus E=\{z:1\le|z|<2\} $$

a "half-open" annulus that includes the inner circle $|z|=1$.

Step 3 · Show this set is not open

Take $z_0=1\in S\setminus E$. For any $\varepsilon>0$, the point $z_\varepsilon=\left(1-\frac\varepsilon2\right)$ satisfies $|z_\varepsilon-z_0|=\frac\varepsilon2<\varepsilon$, so $z_\varepsilon\in B(z_0,\varepsilon)$. But $|z_\varepsilon|=1-\frac\varepsilon2<1$, so $z_\varepsilon\in E$, hence $z_\varepsilon\notin S\setminus E$.

So no disk $B(z_0,\varepsilon)$ is contained in $S\setminus E$ — $z_0$ has no neighborhood inside $S\setminus E$, so $S\setminus E$ is not open.

Step 4 · Why this happens in general

Write $S\setminus E=S\cap E^c$. Since $E$ is open, its complement $E^c$ is closed — so $S\setminus E$ is the intersection of an open set with a closed one, which is generally neither open nor closed. The boundary points of $E$ that happen to lie in $S$ get pulled into $S\setminus E$, and such points always fail the open-set test (any disk around them dips back into $E$).

Step 5 · When does it work out to be open?

$S\setminus E$ is open exactly when $E$ is also closed relative to $S$ (i.e. $E$ contains every one of its limit points that lies in $S$) — for instance when $E=\varnothing$ (giving $S\setminus E=S$), or more generally whenever $E$'s boundary (within $S$) doesn't actually touch $S$.

Verification

Direct check: $z_0=1$ is in $S\setminus E$ (since $1\le|1|=1<2$ ✓), yet points arbitrarily close to $z_0$ with modulus slightly less than $1$ (e.g. $z=0.999$) lie in $E$ and are excluded from $S\setminus E$ — confirming $z_0$ is a boundary point of $S\setminus E$ that belongs to the set but has no open neighborhood inside it.