Proof by contradiction:$E_1 +E_2\doteq E_1\oplus E_2\leftrightarrow \forall x \in E_1+E_2(\exists! e_1 \in E_1,e_2 \in E_2(x=e_1+e_2))$

Question

I must by contradiction:

let $E_1,E_2$ two vector subspacesof $V$, then: $$E_1 + E_2\doteq E_1 \oplus E_2 \leftrightarrow \forall x \in E_1+E_2(\exists! e_1 \in E_1,e_2 \in E_2(x=e_1+e_2))$$

I must show: $$1)E_1 + E_2\doteq E_1 \oplus E_2 \to \forall x \in E_1+E_2(\exists! e_1 \in E_1,e_2 \in E_2(x=e_1+e_2))$$ $$\mbox{and}$$

$$2)\forall x \in E_1+E_2(\exists! e_1 \in E_1,e_2 \in E_2(x=e_1+e_2)) \to E_1 + E_2\doteq E_1 \oplus E_2$$

$2)$ the proof is similar to case $2)$ of CLIC

$1)$ by contradiction I have $e_1,e_1' \in E_1, e_2,e_2' \in E_2(x=e_1+e_2 \wedge x=e_1'+e_2'\wedge e_1\neq e_1' \wedge e_2 \neq e_2')$, I consider $$x-x=e_1'+e_2'-e_1-e_2=e_1'-e_1+e_2'-e_2=0$$ but $e_1'-e_1 \in E_1$ and $e_2'-e_2 \in E_2$ (then $e_1'-e_1+e_2'-e_2 \in E_1 + E_2$) and by hypothesis $E_1+E_2 \doteq E_1 \oplus E_2$ therefore $e_1'-e_1=0 \wedge e_2'-e_2=0$ then $e_1'=e_1 \wedge e_2'=e_2$ and it is an absurd with $e_1\neq e_1' \wedge e_2 \neq e_2'$

It is correct? Thanks in advance!

score 1 · Accepted Answer · answered Feb 26 '14 at 13:49

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I think that you are absolutely right, although this should not be regarded as an answer.

answered Feb 26 '14 at 13:49

Joy-Joy

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okok thanks for your confirmation :) – mle Feb 26 '14 at 13:52
You are welcome. I think you have written out the details of the standard proof of the question. – Joy-Joy Feb 26 '14 at 13:55

Proof by contradiction:$E_1 +E_2\doteq E_1\oplus E_2\leftrightarrow \forall x \in E_1+E_2(\exists! e_1 \in E_1,e_2 \in E_2(x=e_1+e_2))$

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