Dimension of B(T), the set of all bounded functions in a non-empty set T.

Question

I needed to proof:

"For any non-empty set T, the dimension of $\textbf{B}(T)$ is equal to the number of elements of $T$."

My professor gave a tip: use indicator functions. I've managed to find a reasoning, but I think the notation is still flawed. I believe I need to find a basis for the set of all bounded functions; then, the dimension of this basis is the dimension of $\textbf{B}(T)$.

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Proof: Fix $T$ a non-empty set. Let $\textbf{B}(T)$ be the set of all bounded functions in T. For each function (assumed that $f_i: T \rightarrow \mathbb{R}$, for all $i$ in $\mathbb{N}$) we have that $f_i \leq M_i$, $\forall t \in T$.

Consider a singleton $\{x\}$ in T. Define $1_{\{x\}}$ as the indicator function if $t = x$, $\forall t \in T$.

Then the set {$1_{\{x\}}f_i(x)$: ${x} \subseteq T$ and $f_i(x) \leq M_i$, for all $i \in \mathbb{N}$} is a basis for $B(T)$ if T is finite, since it is linearly independent and it is $\supseteq$-minimal in T, since it is constructed with the indicator functions. As such, $dim(B(T))$ = $|T|$.

If T is not finite, we have anyway that $dim(B(T)) \geq |T|$ (since the set T is linearly independent), which forces $dim(B(T)) = \infty.$

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So, is the proof alright? I know it is wrong to use $i \in \mathbb{N}$, so how can I proceed with the notation? Since bounded functions in T are uncountable, how can I write the notation to indicate them? Should I use $i$ in $\mathbb{R}$?

Thanks

Answer 1

It is not true that the dimension of $\mathbf{B}(T)$ equals the cardinality of $T$. For example, let $T$ be the positive integers (for this choice of $T$, the vector space $\mathbf{B}(T)$ is often called $\ell^\infty$). Then $T$ has countable cardinality. However, every basis of $\mathbf{B}(T)$ is uncountable.

To see that every basis of $\ell^\infty$ is uncountable, suppose we had a countable basis $v_1, v_2, \dots$ of $\ell^\infty$. Let $S_n$ denote the span of $v_1, \dots, v_n$. Then $$ \ell^\infty = \bigcup_{n=1}^\infty S_n. $$ However, the equation above violates the Baire Category Theorem, which states that a complete metric space (use the supremum norm on $\ell^\infty$) is not the countable union of closed sets with no interior.

Answer 2

Since the functions $\{\boldsymbol 1_x : x \in T\}$ are linearly independent, you always have $\dim B(T) \ge |T|$. If $|T| = \infty$ this forces $\dim B(T) = \infty$. You get equality for free.

If on the other hand $|T| < \infty$ then in fact $\{\boldsymbol 1_x : x \in T\}$ is a basis because any bounded $f$ on $T$ satisfies $$ f(t) = \sum_{x \in T} f(x) \boldsymbol 1_x(t)$$ for all $t \in T$. Regarding $f$ and $\boldsymbol 1_x$ as elements of $B(T)$ you get $$f = \sum_{x \in T} f(x) \boldsymbol 1_x$$ so that the $\boldsymbol 1_x$'s span $B(T)$. You don't need any special indexing notation: rather index by $x \in T$.

In fact the same summation notation works in case $T$ is infinite, since only one summand $f(x) \boldsymbol 1_x$ is nonvanishing.