Your proof is clear and well-structured, but it could use slight clarification and rigor to ensure that every step logically follows and is fully justified. Here's a review and enhancement of your solution:

Strengths

1. Clarity in Definitions: You clearly defined the height of a rooted tree and the path $P(a,b)$ between two nodes $a, b \in V$. This helps the reader follow your reasoning.
2. Key Relationships: You used the relationships $d(u,x) = d(u,v) + d(v,x)$ and $d(v,y) = d(v,u) + d(u,y)$, which are foundational to the argument.
3. Logical Flow: The proof transitions smoothly between the inequalities involving $h_u$ and $h_v$.

Improvements and Justifications

1. Clarify Symmetry: When concluding $h_v \leq d(v,u) + h_u$, emphasize that $d(u,v) = d(v,u)$ is symmetric because $T$ is a tree.
2. Bound $d(u,v)$: In the final step, where $h_u \leq 2h_v$, explicitly argue why $d(u,v) \leq h_v$. This is essential to ensure the bound holds.
3. Use of Maximum Distance: Reinforce that $d(u,v)$ must be less than or equal to the height of any rooted tree because $u$ and $v$ are part of the same connected structure.

Revised Proof

Let $h_u$ and $h_v$ denote the heights of $(T,u)$ and $(T,v)$, respectively. The height of a rooted tree is the maximum distance from the root to any node in the tree. Thus: $h_u = \max_{x \in V} d(u,x), \quad h_v = \max_{y \in V} d(v,y).$

Since $T$ is a tree, there is a unique path $P(a,b)$ between any two nodes $a, b \in V$, and the distance $d(a,b)$ is defined as the number of edges on this path.

Bounding $h_u$ in Terms of $h_v$

For any node $x \in V$, if $v$ lies on the unique path from $u$ to $x$, then: $d(u,x) = d(u,v) + d(v,x).$ Taking the maximum over all $x \in V$, the height of $(T,u)$ becomes: $h_u = \max_{x \in V} d(u,x) = d(u,v) + \max_{x \in V} d(v,x) = d(u,v) + h_v.$

Bounding $h_v$ in Terms of $h_u$

Similarly, for the farthest node $y \in V$ from $v$ (i.e., $d(v,y) = h_v$): $d(v,y) = d(v,u) + d(u,y).$ Since $d(u,y) \leq h_u$, it follows that: $h_v = d(v,u) + \max_{y \in V} d(u,y) \leq d(v,u) + h_u.$

Combining Inequalities

From the above, we have: $h_u \leq d(u,v) + h_v, \quad h_v \leq d(v,u) + h_u.$ Using $d(u,v) = d(v,u)$, the second inequality becomes: $h_v \leq d(u,v) + h_u.$

Since $d(u,v) \leq h_v$ (as the distance between any two nodes is at most the height of the tree rooted at either node), we substitute $d(u,v) \leq h_v$ into the first inequality: $h_u \leq h_v + h_v = 2h_v.$

Thus, the height of $(T,u)$ is at most twice the height of $(T,v)$.

Conclusion

Your answer is correct and follows logically. The key step is justifying why $d(u,v) \leq h_v$, which you implied but didn’t explicitly explain. Addressing this ensures the argument is airtight.

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Follow-Up Questions

1. Can you provide an example of a tree where $h_u = 2h_v$ to verify the tightness of the bound?
2. How does this proof change if $T$ is not a tree but a general connected graph?
3. What is the geometric interpretation of the relationship between $h_u$ and $h_v$ in terms of tree diameter?
4. Can this proof be extended to trees where edges have weights? How would it change?
5. How would this argument look for a forest (a disjoint union of trees)?

Tip: When proving inequalities, always clearly justify why specific bounds (like $d(u,v) \leq h_v$) hold, even if they seem intuitive.