Linear independence in vector spaces of infinite dimension

Question

Let $V$ be a vector space which has a countable basis. Any set with an uncountable number of elements will hence have to be linearly dependent.

I don't know how to prove the statement above. It is true that if $V$ were finite dimensional, say of dimension $n$, any set of $n+1$ elements would be linearly dependant. How can we prove an analogous statement for vector spaces of infinite dimensions?

Every linearly independent set extends to a basis. This is true for vector spaces of any dimension. — Qiaochu Yuan, Jan 18 '16 at 05:47
When you say "I don't understand the statement above" do you mean not understand the meaning or not see how to prove? It seems unclear right now. — Milo Brandt, Jan 18 '16 at 05:47
I believe he's asking how to prove that in a vector space with a countably infinite basis, that an uncountable set of vectors is linearly dependent. — AnotherPerson, Jan 18 '16 at 05:49
A related question which is sufficient for a proof was asked and answered on MSE here and probably elsewhere on MSE. — André Nicolas, Jan 18 '16 at 05:58

score 4 · Answer 1 · edited Jan 18 '16 at 06:13

4

Let $B$ be a countable basis of $V$. Let $S$ be an uncountable subset of $V$. Suppose $S$ is linearly independent. Then there exists a basis $C$ of $V$ such that $S \subset C$. Since every basis of a vector space has the same cardinal number (see any textbook on abstract algebra that deals with infinite-dimensional vector spaces), $C$ is countable. This is a contradiction.

edited Jan 18 '16 at 06:13

Michael Hardy

1

answered Jan 18 '16 at 05:58

Makoto Kato

42,602

score 4 · Answer 2 · answered Jan 18 '16 at 06:16

Let $\Sigma = \{ S \subset \mathbb{N} | S \text{ is finite} \}$ and note that $\Sigma$ is countable.

Let $\{b_k\}$ be the countable basis.

Suppose $U$ is the uncountable set and $x \in U$. Suppose $x = \sum_{k \in I} \alpha_k b_k$, with $\alpha_k \neq 0$ for all $k \in I$. Define $\phi:U \to \Sigma$ by $\phi(x) = I$, that is, the indices of the support of $x$ in the basis $ \{b_k\}$.

Since $U$ is uncountable, there must be some $I \in \Sigma$ such that $\phi^{-1}(I)$ is uncountable. Then any collection of $|I|+1$ elements of $\phi^{-1}(I)$ will be linearly dependent.

marshal craft · Answer 3 · 2016-01-19T08:26:46.620

To be formal, if $V$ is countable, then there exists a map $ f:V \to A $ such that $ A \subseteq \Bbb N$ and $f$ is surjective. In order for this "set" (lets call it $B$) to be uncountable(a finite set has finite power set or set of all subsets), $V$ must have infinite cardinality and $B \subseteq \mathcal PV$ where $\mathcal PV$ denotes the power set of $V$. Also I am assuming the set $\{\{ v,\{v,m\}\}| v,m \in V\}$ is an acceptable element of $B$.

We see really we have just one infinite countable $V$ to worry about, that is $\Bbb Q$ and the set of elements of $\Bbb Q$ used to construct any real number (infinite basis are Cauchy sequences or rationals). Between any two integers we know there is an uncountable number of real numbers. So here we have an example of a infinite basis witch is countable, yet yields an uncountable infinite vector space.

Additionally we see that the real numbers as an infinite vector space are not linearly independent. They are spanned by the rationals and $dim\text{ }\Bbb Q = dim\text{ }\Bbb R$ when viewed as vector spaces, though the cardinality is certainly not the same.

Linear independence in vector spaces of infinite dimension

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