# For $X$ not necessarily finite there exists a subspace $Y$ of codimension $1$ and $f:X\to\mathbb{K}$ such that $Y=\ker f$.

I need show this exercise:

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## If $$Y\neq 0$$ is a vectorial subspace of $$X$$ so that $$\dim (X/ Y)=1$$ then exist a lineal functional $$f:X\to \mathbb{K}$$ so that $$Y=\ker f$$.

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I already showed this in the case that $$X$$ has a finite dimension (using bases). This is my proof for finite case:

If $$Y$$ is a subspace of $$X$$ of codimension $$1$$ and $$\dim X=n$$, let $$\{y_1, \ldots, y_{n-1}\}$$ be a basis of $$Y$$ and let $$y_n\in X$$ so that $$\{y_1, \ldots, y_n\}$$ is a basis de of $$X$$. Clearly the linear functional $$f :X \to \mathbb{K}$$ so that $$f(y_i)=0$$ for $$i=1, \ldots, n-1$$ and $$f(y_n)=1$$ satisfies that $$Y=\ker f$$.

How can I show my exercise without using finite dimension of X?

By assumption there is a vector $$x_0\in X\setminus Y$$ such that $$X/Y=span\{x_0+Y\}$$. So now define $$f:X\to \mathbb{K}$$ by $$f(x)=\lambda$$ where $$x+Y=\lambda x_0+Y$$. This is clearly a well defined linear functional, and we have $$f(x)=0$$ if and only if $$x+Y=0+Y$$, which happens if and only if $$x\in Y$$.
$$\dim (X/Y)=1$$, therefore, by definition, there is an isomorphism $$\phi:X/Y\to\Bbb K^1$$. Then, you can consider the functional $$f=\phi\circ \pi_Y$$, where $$\pi_Y:X\to X/Y$$ is the quotient map.
• why $\ker(\phi\circ \pi_Y)=Y$? – Luis Prado Sep 26 '20 at 19:55