New York City Mayor Bill de Blasio announced he is backing Hillary Clinton for the 2016 Democratic presidential nomination (video below).
“The candidate who I believe can fundamentally address income inequality effectively, who has the right vision and the right experience to get the job done, is Hillary Clinton,” de Blasio said Oct. 30 on MSNBC’s “Morning Joe.”
De Blasio and Clinton’s relationship is steeped in history as he was her campaign manager when she won a New York Senate seat in 2000. In April, de Blasio declined to endorse Clinton for the Democratic presidential nomination and skipped her official campaign launch on Roosevelt Island, New York, in June, CNN reports.
De Blasio was reportedly waiting until after Clinton addressed her “larger vision” on the issue of income inequality before deciding on whether or not to give her his endorsement.
"I have seen her vision and her platform develop over five months," de Blasio said on the show. "I'm extremely pleased with what she's put on the table but she has a history of fighting on issues that convinces me."
“Hillary Clinton knows how to do it, will get it done and has a progressive platform that speaks to all we need,” de Blasio added.
Clinton issued a statement that referenced de Blasio’s endorsement, while touting the additional mayoral support she has from those across the country.
"I'm honored to receive the support of more than 85 mayors across the country today — because mayors know how to get things done," Clinton said in a statement obtained by The Hill.
"As senator, I worked with mayors from across New York who were committed to finding creative ways to help families get ahead and stay ahead," Clinton added. "As president, I will fight alongside mayors to rebuild and modernize our infrastructure, get middle class income rising and address the epidemic of gun violence."
Other Democratic mayors who have shown their support for Clinton include Chicago Mayor Rahm Emanuel, Philadelphia Mayor Michael Nutter and Mayor Joe Riley of Charleston, South Carolina.