# Suppose X is normed space such that $dim(X) = \infty$, then $dim(B(X,\mathbb{K})) = \infty$

I want to see if my proof is correct. Here $\mathbb{K}$ is the ground field. Suppose that $dim(B(X,\mathbb{K})) = n$ where n is any arbitrarily integer we shall construct $n + 1$ linearly independent vectors in $B(X,\mathbb{K})$. Since X is infinite-dimensional it has n + 1 linearly independent vectors denote them by $x_1,...,x_{n + 1}$

Since X has a hamel basis each $x_i$ can be written as $x_i = \Sigma_i^m e_i\beta_i$ where $\beta_i \in \mathbb{K}$ for each i.

Consider The $\alpha_i : X \rightarrow \mathbb{K}$ defined $\alpha_i(x) = \alpha_i(\Sigma_i^m e_i\beta_i) = \beta_i$. Then one can easily prove that $\alpha_1,...,\alpha_{n + 1}$ are linearly indepedent vectors in $B(X,\mathbb{K})$ and we are done.

## 2 Answers

This is not correct, because you haven't proved your $\alpha_i$ are bounded (and in fact, they probably won't be). To ensure that they're bounded, you're going to have to construct them more carefully instead of just picking any Hamel basis. For instance, you might try using Hahn-Banach.

(There are also various imprecisions in what you've written: you never actually stated what your Hamel basis is, you are using the index $i$ with multiple meanings in a single equation, and your argument appears to not actually make any use of the $x_1,\dots,x_{n+1}$ you chose at the beginning. But even if you fixed all of these, nothing along the lines of what you're writing could work because you are doing nothing that will ensure your functionals are bounded.)

• Yes I see. I will think about this more carefully. – user111750 Nov 2 '16 at 20:53
• Yeah your right so I proved it using Hahn Banach. I guess a corollary that comes from Hahn Banach theorem. – user111750 Nov 3 '16 at 1:07

So I was able to prove. Maybe this will help people. So using Hahn Banach theorem we can prove the following theorem. Suppose $dim(B(X,\mathbb{K})) = n$

Suppose X is normed space and Y is closed subspace of Y. Then there exists $f \in B(X,\mathbb{K})$ such that $f(x) = 0\ \forall x \in Y$ and f(z) = 1 for $z \notin Y$.

Consider $x_1 \in X$, then consider $Y_1 = span(x_1)$ and suppose that $x_2 \notin span(x_1)$. Then by theorem above we can construct $f_1$ that kills $Y_1$ and is 1 otherwise. Continuing this way we can construct n + 1 linearly indepedent vectors, so $dim(X,\mathbb{K}) > n$ (this happens for all n). So we get our conclusion.