Let $C=\{(0,0,0,0,0),(1,1,1,0,0),(0,0,1,1,1),(1,1,0,1,1)\}\in\mathbb{F}_2$. Show $C$ is $1$ error correcting.
Definition: a code $C\subseteq\mathbb{F}^n$ is t error correcting , if for any two distinct codewords $x\neq y,x,y\in C$, and any two vectors $\alpha,\beta\in\mathbb{F}^n$ with $wt(\alpha),wt(\beta)\leq t$, we have $x+\alpha\neq y+\beta$
I think I can do this by contradiction.
Attempt: Assume that $C$ is not $1$ error correcting, there exist two codewords $x, y \in C$ and two vectors $a,b\in\mathbb{F}^n$ with $wt(a),wt(b) \leq 1$ such that $x+a=y+b$. Note that $C$ is a $[5,2]$ linear code with basis $\{(1,1,1,0,0),(0,0,1,1,1)\}$, the minimum distance of $C$ is $3$.
At this step, I notice my proof is wrong because I don't see how to receive a contradiction. Can someone give me a hint or suggestion to do this problem? Thanks.
Since $x+\alpha \neq y+\beta$, we can show that $x-y \neq \beta-\alpha$. As both $\alpha$ and $\beta$ have weights that are less than or equal to $t$, $\beta-\alpha$ must have a weight that is less than or equal to $2t$. Thus, we just need to show that the weight of $x-y$ is at least $2t+1$ for all $x, y \in C$.
In our case, $t=1$, so we need to show that the weight of $x-y$ is at least $3$. However, from the work that you already did, it seems that you already proved this. Thus, by proving that $x-y$ has a weight of at least $3$ for all $x, y \in C$, we have shown that $C$ is 1-error correcting.
In my opinion, it's much easier to prove that a code is error correcting by looking at the minimum distance between codes rather than trying to show a contradiction, especially if the code is very small like this one. If proof by contradiction works for you better than this method, then go for it, but otherwise, try the minimum distance method.