Continous functions and supremum

Question

this might be a very trivial questions that can be answered by pointing to a theorem or lemma, but I have trouble remembering something clearly. Give a continuous function $f(x)$ with $x\in \mathbb{R}^n$. Why is (or is): $$ \sup_{x\in \mathbb{R}^n\backslash\{0\}}f(x)=\sup_{x\in \mathbb{R}^n}f(x) $$ Some similar equation came up in a class about a continuous likelihood function (for a Gaussian distribution) where the supremum of the likelihood function over $\beta\in \mathbb{R}^n$ was equated to the supremum of the likelihood function with $\beta\in \mathbb{R}^n, K\beta\neq c$ where $K \in \mathbb{R}^{k\times n}$ is a contrast matrix and $c\in \mathbb{R}^k$.

Answer 1

It's because $0 \in \mathbb{R}^n$ is a limit point of $\mathbb{R^n}\backslash 0$. To prove the claim, note that there is nothing to prove if the supremum of $f$ is not attained at $0$. Suppose then that $\sup_{x \in \mathbb{R}^n}f(x) = f(0)$. Continuity of the function implies that $$ \lim_{x\to 0}f(x) = f(0).$$ As such, for all $\epsilon > 0$ there exists $x \neq 0$ such that $f(0) < f(x) + \epsilon$. This implies that $$\sup_{x \in \mathbb{R}^n \backslash 0}f(x) > f(0)-\epsilon$$ for all $\epsilon$. Hence the two suprema are equal.