<｜begin▁of▁sentence｜>### Instruction:
Assuming you have a set of elements [1, 2, 3, 4, 5], and you wish to perform a union operation to connect 1 with 2 and also to connect 3 with 4, while keeping 5 as a separate set. How can you use networkx's UnionFind structure to do this, and afterward determine if 1 and 3 are in the same set?
### Response:
import networkx as nx

# Create a UnionFind object
uf = nx.utils.UnionFind()

# Elements to be unioned
elements = [1, 2, 3, 4, 5]

# Add elements
for element in elements:
    uf[element]

# Union elements
uf.union(1, 2)
uf.union(3, 4)

# Find the roots
root_1_2 = uf[1]
root_3_4 = uf[3]

# Print roots to show union
print('Root of 1:', root_1_2)
print('Root of 2:', uf[2])
print('Root of 3:', root_3_4)
print('Root of 4:', uf[4])

# Check if 2 elements are connected (have the same root)
is_connected_1_3 = uf.connected(1, 3)
print('Are 1 and 3 connected?:', is_connected_1_3)
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