proof: $r \in \langle S\setminus \{r\}\rangle \leftrightarrow \langle S\rangle = \langle S\setminus \{r\} \rangle$

Question

Theorem: Let be $V$ a $K-$vector space, and $ r \in S \subseteq V$: $$r \in \langle S\setminus \{r\}\rangle \leftrightarrow \langle S\rangle = \langle S\setminus \{r\} \rangle$$ Proof:$$ "\to"$$ $"\subseteq"$ -- I prove that $S \subseteq \langle S\setminus \{r\} \rangle$: $$ \begin{align} x \in S &\to x \in \{r\} \cup S \setminus \{r\} \\ &\to x = r \vee x \in S \setminus \{r\} \\ &\to x \in \langle S\setminus \{r\}\rangle \text{ by Iphotesis } \vee \\ &\vee x \in \langle S\setminus \{r\}\rangle \text{ because }S\setminus \{r\} \subseteq \langle S\setminus \{r\}\rangle \end{align}$$ Now it is simple, in fact $$S \subseteq \langle S\setminus \{r\}\rangle \to \langle S\rangle \subseteq \langle \langle S\setminus \{r\}\rangle \rangle= \langle S\setminus \{r\}\rangle \to \langle S \rangle \subseteq \langle S\setminus \{r\}\rangle$$ $"\supseteq"$-- Naturally $S\setminus \{r\}\subseteq S$ with $r \in S$, therefore: $$ \langle S\setminus \{r\}\rangle \subseteq \langle S \rangle$$ $$"\leftarrow"$$ We have by Iphotesis that $r \in S$, therefore: $$r \in S \subseteq \langle S\rangle=\langle S\setminus \{r\}\rangle \to r \in \langle S\setminus \{r\}\rangle$$

Is it correct?

Answer 1

The part of the proof showing that $S \subseteq \langle S \setminus \{r\}\rangle$ can be written slightly more elegantly as follows:

Clearly one has $S \setminus \{ r\} \subseteq \langle S \setminus \{r\}\rangle$ Furthermore, $\{r\} \subseteq \langle S \setminus \{r\}\rangle$ by assumption. Hence $$S = (S \setminus \{r\}) \cup \{r\} \subseteq \langle S \setminus \{r\}\rangle\,.$$ Otherwise the proof looks good.