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Denote $[S]$ as the span of $S$.
Let $S$ be a subset of a vector space $V$ and $\alpha \in V$. Then $\alpha \in [S]$ if and only if $[S]=[S\backslash \{\alpha\}]$.

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Incorrect as written. If $V=R^2$ and $S=\{(1,0),(0,1)\}$ and $\alpha = \{(1,0)\}$, $\alpha \in V$ but $[S]=R^2=V$ and $[S-\{\alpha\}]=[\{(0,1)\}] \neq R^2=V$. –  kigen Dec 11 '12 at 7:24
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You certainly don't want $\alpha\in V$ as hypothesis and the showing "$\alpha\in V$ if and only if ...". You probably mean $\alpha\in[S]$ before the "if and only if", but even then it does not make real sense; please say what you want to ask. –  Marc van Leeuwen Dec 11 '12 at 7:26
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What does $S-\{\alpha\}$ mean? (a) $S\setminus\{\alpha\}$ or (b) $\{v-\alpha: v\in S\}$? –  user1551 Dec 11 '12 at 7:34
    
My bad.I edited the question above sir. –  Philip Benj Marcoby Eragon Dec 11 '12 at 8:02
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up vote 1 down vote accepted

Maybe you meant to conclude "$\alpha\in[S]$ if and only if $[S]=[S\cup\{\alpha\}]$", which is correct. But you cannot expect people to read your mind, so please state more carefully.

For the edited question "$\alpha \in V$. Then $\alpha \in [S]$ if and only if $[S]=[S\backslash \{\alpha\}]$" it is certainly incorrect; for one thing if $\alpha\notin S$ then $[S]=[S\backslash \{\alpha\}]$ holds trivially, regardless of whether $\alpha\in [S]$, while if $\alpha\in S$ then $\alpha \in [S]$ holds trivially, regardless of whether $[S]=[S\backslash \{\alpha\}]$.

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My bad.I edited the question above sir. –  Philip Benj Marcoby Eragon Dec 11 '12 at 8:02
    
Thank you sir. I appreciate this post. –  Philip Benj Marcoby Eragon Dec 11 '12 at 8:19
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