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About: Hornellsville, New York     Goto   Sponge   NotDistinct   Permalink

An Entity of Type : umbel-rc:Village, within Data Space : dbpedia.demo.openlinksw.com associated with source document(s)

Type: Thingplacedbo:Locationschema:PlaceTownHuman settlementpopulated placeyago:WikicatCitiesInNewYorkyago:WikicatTownsInNewYorkyago:WikicatTownsInSteubenCounty,NewYorkgeo:SpatialThingyago:AdministrativeDistrict108491826yago:City108524735yago:District108552138yago:GeographicalArea108574314yago:Location100027167yago:Municipality108626283yago:Object100002684yago:PhysicalEntity100001930yago:Region108630985yago:YagoGeoEntityyago:YagoLegalActorGeoyago:YagoPermanentlyLocatedEntitysettlementtownyago:Site108651247yago:Town108665504yago:Tract108673395yago:UrbanArea108675967yago:WikicatPopulatedPlacesEstablishedIn1790yago:WikicatPopulatedPlacesInSteubenCounty,NewYorkumbel-rc:Location_Underspecifiedumbel-rc:PopulatedPlaceumbel-rc:Townumbel-rc:Village New Facet based on Instances of this Class

Hornellsville is a town in Steuben County, New York, United States. The population, not counting the city of Hornell, was 4,039 at the 2020 census. The name is taken from a prominent pioneer family. The Town of Hornellsville is at the western border of the county, and surrounds the city of Hornell. Until 1906, the city of Hornell was named Hornellsville. The name Hornellsville is only used in real estate or legal contexts, but rarely in conversation. This is because Hornell is much more frequently mentioned, and its boundaries are quite different.