# A vector space written as a finite union of proper subspaces

We raise here the following question: “can a vector space $$E$$ be written as a finite union of proper subspaces”?

Let’s consider the simplest case, i.e. writing $$E= V_1 \cup V_2$$ as a union of two proper subspaces. By hypothesis, one can find two non-zero vectors $$v_1,v_2$$ belonging respectively to $$V_1 \setminus V_2$$ and $$V_2 \setminus V_1$$. The relation $$v_1+v_2 \in V_1$$ leads to the contradiction $$v_2 = (v_1+v_2)-v_1 \in V_1$$ while supposing $$v_1+v_2 \in V_2$$ leads to the contradiction $$v_1 = (v_1+v_2)-v_2 \in V_2$$. Therefore, a vector space can never be written as a union of two proper subspaces.

We now analyze if a vector space can be written as a union of $$n \ge 3$$ proper subspaces. We’ll see that it is impossible when $$E$$ is a vector space over an infinite field. But we’ll describe a counterexample of a vector space over the finite field $$\mathbb{Z}_2$$ written as a union of three proper subspaces. Continue reading A vector space written as a finite union of proper subspaces

# Counterexamples on real sequences (part 1)

I will come back later on with more complex cases. Unless otherwise stated, $$(u_n)_{n \in \mathbb{N}}$$ and $$(v_n)_{n \in \mathbb{N}}$$ are two real sequences. Continue reading Counterexamples on real sequences (part 1)

# A solution of a differential equation not exploding in finite time

In this post, I mention that Peano existence theorem is valid for finite dimensional vector spaces, but not for Banach spaces of infinite dimension. I highlight here a second property of ordinary differential equations which is valid for finite dimensional vector spaces but not for infinite dimensional Banach spaces. Continue reading A solution of a differential equation not exploding in finite time

# They were killed but their freedom of speech remains

For a continuous function $$f: X \mapsto Y$$, the preimage $$f^{-1}(V)$$ of every open set $$V \subseteq Y$$ is an open set which is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in $$Y$$ are closed in $$X$$. However, a continuous function might not be an open map or a closed map as we prove in following counterexamples. Continue reading Continuous maps that are not closed or not open