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.


Your proof is generally fine; I can follow your logic. If we really want to be nitpicky:


*

*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."

