Eulerian circuit with no isolated vertex is connected

This is my first question (ever), and I am pretty new to math. So I ask for patience and understanding in advance.

So this is the proof I came up with:

Consider $G = (V,E).$ By definition of Eulerian circuit, $$\exists\ W: v_0, v_1,\cdots, v_k=v_o$$ And there is no isolated vertex $$V = \{v_0, v_1,\cdots,v_k \}$$
Take $v_i, v_j \in V$. Consider $j = i + k$ where $k \in \mathbb{N}$. Then, $v_j, v_{j+1}, \cdots, v_{j+k} = v_i$ is a walk. Hence, $\exists$ a path between $v_i$ and $v_j$. Hence connected.

• So what is your question? Are you asking whether your proof is correct? – Joao Oct 28 '14 at 3:48
• Yes. I feel a little uncomfortable proving graph theorems so I wanted some feed-backs (and perhaps some assurance). – Henri L Oct 28 '14 at 4:15

• You should probably mention that we are assuming that $G = (V,E)$ has an Eulerian circuit and has no isolated vertices.
• You should say: "Since $G$ has no isolated vertices, we know that $V = \{v_0, \ldots, v_k\}$."
• When you take $v_i, v_j \in V$, you should say that they are distinct and that we are assuming, without loss of generality, that $i < j$. This guarantees that $k$ really is a natural number.
• When you explicitly specify the walk between $v_i$ and $v_j$, you seem to have mixed up $i$ with $j$.
• It's usually better to use complete English sentences in proofs. Avoid unnecessary logic symbols. Finish your proof with something like: "Hence, there exists a path between $v_i$ and $v_j$ so that $G$ is connected, as desired."