If $x$ is a local minimum of $f$ restricted to an open subset, why can we conclude it is a minimum of $f$ as well?

Question

Let $(E,\tau)$ be a topological space and $f:E\to\mathbb R$. We say that $x\in E$ is a local minimum of $f$ if $$f(x)\le f(y)\;\;\;\text{for all }y\in N\tag1$$ for some open neighborhood $N$ of $x$.

Let $\Omega\subseteq E$ and $x$ be a local minimum of $\left.f\right|_\Omega$. If $\Omega$ is open, why can we conclude that $x$ is a local miminum of $f$?

Answer 1

If $x \in \Omega$ is a local minimum of $\left.f\right|_\Omega$, then there is an open neigborhood $N \subset \Omega$ of $x$ such that $f(x) \le f(y) $ for all $y \in N.$

Since $\Omega$ is open, $N$ is open in $E$, and the result follows.