I have the following question and I'm not sure my proof is correct/approached correctly:

Let $V$ be a $n$-dimensional vector space, let $U_i \subset V$ be subspaces of V for $i = 1,2,\dots,r$ where $$U_1 \subset U_2 \subset \dots \subset U_r$$ If $r>n+1$ then there exists an $i<r$ for which $U_i = U_{i+1}$

I was thinking something along the lines of: the subsets are strict thus 'moving' from $U_i$ to $U_{i+1}$ increases the dimension by one, which means, when reaching $U_r$, your dimension is greater than $n+1$ but this is not possible since $U_r$ is a subspace of $V$ and thus has dimension at most $n$.

Is this a good approach to this proof? I'm not sure because of the strict inclusions: is it true for every case that $dim(U_i) < dim(U_{i+1})$?

I was maybe thinking of proving this via induction, but I'm not sure how that would work.

Thanks in advance.

EDIT: Updated subset-strictness

  • 2
    $\begingroup$ Yes, it's true (if $U_i\varsubsetneq U_{i+1}$). So you can deduce $\dim U_r>\dim U_i+r-1$. $\endgroup$
    – Bernard
    Commented Jan 24, 2017 at 14:11

3 Answers 3


Your proof is (essentially) correct.

In the second line of the paragraph you want

increases the dimension by at least one

For example, consider $$ \{0\} \subseteq \text{$x$-$y$-plane} \subseteq \ldots $$ in $\mathbb{R}^3$.

Your instructor might want you to prove that claim about dimensions, or might be willing to take it as known, depending on what you've done in class.

You don't need induction.

Note: although the question you're asking is clear, the statement

$$U_1 \subsetneq U_2 \subsetneq \dots \subsetneq U_r$$ If $r>n+1$ ...

isn't the right way to phrase it. The strict inequalities should be weak $\subseteq$. Then you're to show that at least one of them is an equality.

  • $\begingroup$ Could you clarify why this is 'at least one'? $\endgroup$ Commented Jan 24, 2017 at 14:20
  • $\begingroup$ @RobinHaveneers See my edits. $\endgroup$ Commented Jan 24, 2017 at 14:35
  • $\begingroup$ Well, the actual question just uses $\subset$, but I believe it is used with the assumption of strictness in my book.. $\endgroup$ Commented Jan 24, 2017 at 15:13
  • 1
    $\begingroup$ Nope, I'm wrong, it's used as 'can be equal but isn't necessarily so'. So your note is the correct assumption. Thank you very much $\endgroup$ Commented Jan 24, 2017 at 15:15

Hints following your own work:

=== Since $\;U_k\subsetneq U_{k+1}\;$ , we get that always $\;\dim U_{k+1}\ge \dim U_k+1\;$ , and from here $\;\dim U_r\ge n\;$ ..

=== If $\;W\le V\;$ are finite dimensional linear spaces and $\;\dim W=\dim V\;$ , then $\;W=V\;$ .


Your idea is very good, although not well expressed; but also the statement is not well written. Let's state it properly.

Let $V$ be a $n$-dimensional vector space, let $U_i \subseteq V$ be subspaces of $V$ for $i = 1,2,\dots,r$ where $$U_1 \subseteq U_2 \subseteq \dots \subseteq U_r$$ If $r>n+1$ then there exists an $i<r$ for which $U_i = U_{i+1}=\dots=U_r$

Let's prove, by induction on $r$, that, given a sequence of subspaces $$ U_1\subsetneq U_2\subsetneq\dots\subsetneq U_r $$ then $\dim U_r\ge r-1$. We only use the fact that $U\subsetneq U'$ implies $\dim U'\ge 1+\dim U$.

If $r=1$, this is trivial. Assume it holds for $r-1$. Then, by induction hypothesis, $\dim U_{r-1}\ge r-2$ and so $\dim U_r\ge (r-2)+1=r-1$.

So, if we have $$ U_1\subseteq U_2\subseteq\dots\subseteq U_r $$ and $r>n+1$, the above result proves that not all inclusions are proper.

Let $j$ be the maximum index such that $U_{j}\subsetneq U_{j+1}$ (or $j=0$ if all inclusions are equalities): by the result above, $j<r-1$. Now, for $i=j+1$, $U_i=U_{i+1}=\dots=U_r$.


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