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Let $X \neq \{0\}$ be a normed vector space, $C \subset X $ a convex subset. Is it true that $C$ is the intersection of some half-spaces in $X$?

So I think that the statement is true, but I find it hard to solve it formally. I need help to write it down.

So I've started from bringing back the definition of half-spaces.

For some $\phi \in X^\# \setminus \{0\}$ and $g \in \mathbb R$ we define half space as: $\phi^{-1}((g,+ \infty))$.

And the definition of convex set:

For all $x,y \in C $ and $t \in(0,1) $ we have $t x + (1-t)y \in C$.

I can't connect this information into one to get the thesis. Please help me

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    what is $X^{\#}$ ? – 2017-01-22
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    it is the set of linear functional from $X$ to $\mathbb R$ – 2017-01-22
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    What happens if $C=X$? – 2017-01-22
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    do you need the number of half spaces to be finite? – 2017-01-22
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    @JorgeFernándezHidalgo In this case it is not true in general. See the disk. – 2017-01-22
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    I think I don't need to – 2017-01-22
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    which disk?${}{}{}$ – 2017-01-22
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    On the plane, of course. This is the intersection of uncountably many half-spaces. – 2017-01-22
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    @szw1710: It is sufficient to take countably many half-spaces. – 2017-01-22

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The statement is not true. Consider the following convex subset of $\mathbb{R}^2$: $$C = \{ (x,y) : 0\leq x < 1, 0 \leq y \leq 1\} \cup \{(1,0)\}$$ This is essentially a square without its right boundary, but including the right lower corner point. It is impossible to "shape" the right side with half spaces. The best half spaces you can use are those that "chop off" the line (and everything to the right of it) through the points $(1,1)$ and $(1+1/n,0)$ for $n=1,2,\ldots$, but these do not "chop off" the right boundary other than the point $(1,1)$.

The statement is true for closed convex sets. A reference using closed half spaces is Theorem 11.5 in the book Convex Analysis by R.T. Rockafellar. If you'd like to use open half spaces, just recall that a closed half space is the intersection of infinitely many open half spaces.

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    An open half-space that includes (1,0) but not (1,1/2) cannot cover C...............+1. – 2017-01-24