Talk:Continuous Image of Compact Space is Compact/Corollary 3

From ProofWiki
Jump to navigation Jump to search

Since $f[S]$ compact, then $f[S]$ is closed, hence $f[S]=\overline{f[S]}$. Let $U\ni x$ be an open neighborhood of $\alpha\,\colon=\sup f[S]$. Then there is an $r>0$ such that $B_r(\alpha)=(\alpha-r,\alpha+r)\subseteq U$. From Characterizing Property of Supremum of Subset of Real Numbers[1], there is a $y\in f[S]$ such that $\alpha-r<y\leq\alpha$, thus $\alpha$ is an adherent point of $f[S]$. As a point is adherent to $f[S]$ if and only if it is in $\overline{f[S]}$, then $\alpha\in \overline{f[S]}=f[S]$.

Alternatively, if $\alpha\,\colon=\sup f[S]$ is not in $f[S]$, then by definition of an open set in a metric space, there is an $r>0$ such that $B_r(\alpha)=(\alpha-r,\alpha+r)\subseteq\mathbb{R}\setminus f[S]$. By the characterization of the supremum in the real numbers, there is a $y\in f[S]$ such that $\alpha-r<y\leq\alpha$. Hence $y\in(\mathbb{R}\setminus f[S])\cap f[S]=\emptyset$ which is a contradiction.

  1. Characterizing Property of Supremum of Subset of Real Numbers [[1]]